def merge_at(s, i):
    s[i] += s[i+1]
    del s[i+1]
    return s[i]

def merge_force_collapse(s):
    cost = 0
    while len(s) > 1:
        n = len(s) - 2
        if n > 0 and s[n-1] < s[n+1]:
            n -= 1
        cost += merge_at(s, n)
    return cost

def sort(runs, merge_collapse):
    stack = []
    maxstack = 0
    cost = 0
    for x in runs:
        stack.append(x)
        maxstack = max(maxstack, len(stack))
        cost += merge_collapse(stack)
    cost += merge_force_collapse(stack)
    return cost, maxstack

def timsort(s):
    cost = 0
    while len(s) > 1:
        n = len(s) - 2
        if ((n > 0 and s[n-1] <= s[n] + s[n+1]) or
            (n > 1 and s[n-2] <= s[n-1] + s[n])):
            if s[n-1] < s[n+1]:
                n -= 1
            cost += merge_at(s, n)
        elif s[n] <= s[n+1]:
            cost += merge_at(s, n)
        else:
            break
    return cost

def twomerge(s):
    cost = 0
    while len(s) > 1:
        n = len(s) - 2
        if s[n] < 2 * s[n+1]:
            if n > 0 and s[n-1] < s[n+1]:
                n -= 1
            cost += merge_at(s, n)
        else:
            break
    if len(s) > 1:
        assert s[-2] >= 2 * s[-1]
        if len(s) > 2:
            assert s[-3] >= 2 * s[-2]
    return cost

# Minimal cost across all possible ways of merging runs.
# Takes time cubic in len(runs).
def ideal(runs):
    runs = list(runs)
    n = len(runs)
    # c[i, w] is minimal cost of merging slice runs[i : i+w].
    c = {(i, 1) : 0 for i in range(n)}
    prefixsum = runs[0]
    for width in range(2, n+1):
        prefixsum += runs[width - 1]
        total = prefixsum
        for start in range(n - width + 1):
            c[start, width] = total + min(c[start, w] + c[start + w, width - w]
                                          for w in range(1, width))
            if start + width < n:
                total += runs[start + width] - runs[start]
    return c[0, n]

def greedy(runs):
    s = list(runs)
    cost = 0
    while len(s) > 1:
        i = min(range(len(s) - 1), key=lambda i: s[i] + s[i+1])
        cost += merge_at(s, i)
    return cost

# Bad sequence of run lengths for timsort, summing to `n`.
def rtim(n):
    if n <= 3:
        return [n]
    n2 = n >> 1
    return rtim(n2) + rtim(n2 - 1) + [(n & 1) + 1]

def power(s1, n1, n2, n):
    assert s1 >= 0
    assert n1 >= 1 and n2 >= 1
    assert s1 + n1 + n2 <= n
    # a = s1 + n1/2
    # b = s1 + n1 + n2/2 = a + (n1 + n2)/2
    a = 2*s1 + n1       # 2*a
    b = a + n1 + n2     # 2*b
    # Array length has d bits.  Max power is d:
    #     b/n - a/n = (b-a)/n = (n1 + n2)/2/n >= 2/2/n = 1/n > 1/2**d
    # So at worst b/n and a/n differ in bit 1/2**d.
    # a and b have <= d+1 bits. Shift left by d-1 and divide by 2n =
    # shift left by d-2 and divide by n.  Result is d - bit length of
    # xor.  After the shift, the numerator has at most d+1 + d-2 = 2*d-1
    # bits. Any value of d >= n.bit_length() can be used.
    d = n.bit_length()  # or larger; smaller can fail
    a = (a << (d-2)) // n
    b = (b << (d-2)) // n
    return d - (a ^ b).bit_length()

# Like power but move n up to a power of 2.
def powercheat(s1, n1, n2, n):
    assert s1 >= 0
    assert n1 >= 1 and n2 >= 1
    assert s1 + n1 + n2 <= n
    a = 2*s1 + n1       # 2*a
    b = a + n1 + n2     # 2*b
    d = n.bit_length()  # or larger; smaller can fail
    a >>= 2
    b >>= 2
    return d - (a ^ b).bit_length()

def powerloop(s1, n1, n2, n):
    assert s1 >= 0
    assert n1 >= 1 and n2 >= 1
    assert s1 + n1 + n2 <= n
    # a = s1 + n1/2
    # b = s1 + n1 + n2/2 = a + (n1 + n2)/2
    a = 2*s1 + n1       # 2*a
    b = a + n1 + n2     # 2*b
    L = 0
    while True:
        L += 1
        if a >= n:
            assert b >= n
            a -= n
            b -= n
        elif b >= n:
            break
        assert a < b < n
        a <<= 1
        b <<= 1
    return L

from math import frexp
T53 = float(2**53)

# Using frexp works fine for small n, but flakes out quickly as n grows.
# To get it to flake out even faster, instead of dividing by n, multiply
# by 1.0/n.
def power_frexp(s1, n1, n2, n):
    assert s1 >= 0
    assert n1 >= 1 and n2 >= 1
    assert s1 + n1 + n2 <= n
    # a = s1 + n1/2
    # b = s1 + n1 + n2/2 = a + (n1 + n2)/2
    a = s1 + n1 * 0.5
    b = a + (n1 + n2) * 0.5
    m1, e1 = frexp(a / n)
    m2, e2 = frexp(b / n)
    if e1 != e2:
        return 1 - max(e1, e2)
    m1 = int(m1 * T53)
    m2 = int(m2 * T53)
    x = m1 ^ m2
    assert x
    return 54 - x.bit_length() - e1

if 0:
    for N in range(2, 10001):
        print(N)
        for s1 in range(N-1):
            # 2 <= sum of lengths <= N - s1
            for lensum in range(2, N - s1 + 1):
                for n1 in range(1, lensum):
                    n2 = lensum - n1
                    pold = powerloop(s1, n1, n2, N)
                    pnew = powercheat(s1, n1, n2, N)
                    if pold != pnew:
                        print("OUCH", N, pold, pnew, s1, n1, n2)
                        assert False

if 0:
    from random import randrange
    count = 0
    LO = 1 << 60
    HI = LO << 1
    while True:
        count += 1
        N = randrange(LO, HI)
        s1 = randrange(N-1)
        for lensum in 2, N - s1, randrange(3, N - s1):
            for n1 in 1, lensum - 1, randrange(1, lensum):
                n2 = lensum - n1
                pold = powerloop(s1, n1, n2, N)
                pnew = powercheat(s1, n1, n2, N)
                if pold != pnew:
                    print("OUCH", N, pold, pnew, s1, n1, n2)
                    assert False
        if count % 100_000 == 0:
            print(count)

def powersort(runs):
    cost = 0
    maxstack = 0
    s = []
    runs = list(runs)
    n = sum(runs)
    s1, n1 = 0, runs[0]
    for i in range(1, len(runs)):
        s2, n2 = s1 + n1, runs[i]
        p = power(s1, n1, n2, n)
        #p = powercheat(s1, n1, n2, n)
        while s and s[-1][-1] > p:
            s0, n0, _ = s.pop()
            assert s0 + n0 == s1
            n1 += n0
            cost += n1
            s1 = s0
        assert s1 + n1 == s2
        if s:
            assert s[-1][-1] < p  # never equal!
        s.append((s1, n1, p))
        maxstack = max(maxstack, len(s))
        s1, n1 = s2, n2
    while s:
        s0, n0, _ = s.pop()
        assert s0 + n0 == s1
        n1 += n0
        cost += n1
        s1 = s0
    assert (s1, n1) == (0, n), (s1, n1, 0, n)
    return cost, maxstack

def show(runs):
    runs = list(runs)
    n = sum(runs)
    s1, n1 = 0, runs[0]
    for i in range(1, len(runs)):
        s2, n2 = s1 + n1, runs[i]
        p = power(s1, n1, n2, n)
        print(f"{n1} <{p}>")
        s1, n1 = s2, n2
    print(n1)

if 1:
    from random import random, randrange

    print("all the same") # identical stats
    runs = [32] * 1000
    print(sort(runs, timsort))
    print(sort(runs, twomerge))
    print(powersort(runs))

    print("ascending")  # timsort a little better
    runs = range(1, 2000)
    print(sort(runs, timsort))
    print(sort(runs, twomerge))
    print(powersort(runs))

    print("descending") # twomerge significantly better
    print(sort(reversed(runs), timsort))
    print(sort(reversed(runs), twomerge))
    print(powersort(reversed(runs)))

    print("bad timsort case")
    runs = rtim(100000)
    print(sort(runs, timsort))
    print(sort(runs, twomerge))
    print(powersort(runs))

    print("bad powersort case")
    n = 1000000
    runs = (n-1, 1, 1, n)
    print(sort(runs, timsort))
    print(sort(runs, twomerge))
    print(powersort(runs))

    # "Random" run-length distributions are largely irrelevant, since on
    # randomly ordered input the actual sort is most likely to _force_
    # (via local binary insertion sorts) all runs to length `minrun`.
    # Nevertheless ... there's no clear overall winner in this
    # particular made-up distribution.  On some runs timsort "wins" in
    # the end, on others twomerge, but the total costs are typically
    # within 2% of each other.
    # Later:  powersort almost always dominates both.
    print("randomized trials")
    totals = [0, 0, 0]
    for trial in range(20):
        runs = []
        for i in range(10000):
            switch = random()
            if switch < 0.80:
                x = randrange(1, 100)
            else:
                x = randrange(1000, 10000)
            runs.append(x)
        for i, which in enumerate([timsort, twomerge]):
            cost, depth = sort(runs, which)
            print(f"{which.__name__:8s}", cost, depth)
            totals[i] += cost
        #cost = ideal(runs)
        #print("   ideal", cost)
        #totals[2] += cost
        #cost = greedy(runs)
        #print("  greedy", cost)
        #totals[3] += cost
        cost, depth = powersort(runs)
        print("power   ", cost, depth)
        totals[2] += cost
        print("sofar", totals, "timsort excess",
              f"{(totals[0] - totals[1]) / totals[1]:.2%}")

if 0:
    def p(xs):
        def inner(i):
            if i == n:
                yield xs
                return
            orig = xs[i]
            for j in range(i, n):
                xs[i] = xs[j]
                xs[j] = orig
                yield from inner(i+1)
                xs[j] = xs[i]
            xs[i] = orig
        n = len(xs)
        return inner(0)
    
if 1:
    from itertools import product
    worst = 0.0
    for t in product(range(1, 20), repeat=4):
        g = ideal(t)
        b, ignore = powersort(t)
        assert g <= b
        if g != b:
            ratio = b / g
            if ratio > worst:
                print(t, b, g, ratio)
                worst = ratio
    print("done")
    # n-1, 1, 1, n  # sum = 2n + 1
    # best = 2 + n+1 + 2n+1 = 3n + 4
    # powersort = 4n + 2
    
    # a = (n-1)/2 / (2*n + 1) = (n-1)/(4n+2)
    # b = (n - 1 + 1/2) / (2*n + 1) = (2n-1)/(4n+2)
    # so (0, n-1, 1) has power 2

    # then (n-1, 1, 1)
    # a = n-1 + 1/2 = n - 1/2; (2n-1)/(4n+2)
    # b = n + 1/2l             (2n+1)/(4n+2)
    # so has power 1, and n-1 is merged with 1 first