Problem:
Given is an integer sequence of length N+1: A_0, A_1, A_2, \ldots, A_N. Is there a binary tree of depth N such that, for each d = 0, 1, \ldots, N, there are exactly A_d leaves at depth d? If such a tree exists, print the maximum possible number of vertices in such a tree; otherwise, print -1.

Incorrect submission (verdict: Wrong Answer):
import itertools
from collections import deque,defaultdict,Counter
from itertools import accumulate
import bisect
from heapq import heappop,heappush,heapify
import math
from copy import deepcopy
import queue
#import numpy as np

Mod = 1000000007
fact = [1, 1]
factinv = [1, 1]
inv = [0, 1] 
for i in range(2, 10**5 + 1):
    fact.append((fact[-1] * i) % Mod)
    inv.append((-inv[Mod % i] * (Mod // i)) % Mod)
    factinv.append((factinv[-1] * inv[-1]) % Mod)
    
def cmb(n, r, p):
    if (r < 0) or (n < r):
        return 0
    r = min(r, n - r)
    return fact[n] * factinv[r] * factinv[n - r] % p
    
def sieve_of_eratosthenes(n):
    if not isinstance(n,int):
        raise TypeError("n is not int")
    if n<2:
        raise ValueError("n is not effective")
    prime = [1]*(n+1)
    for i in range(2,int(math.sqrt(n))+1):
        if prime[i] == 1:
            for j in range(2*i,n+1):
                if j%i == 0:
                    prime[j] = 0
    res = []
    for i in range(2,n+1):
        if prime[i] == 1:
            res.append(i)
    return res

 
class UnionFind:
    def __init__(self,n):
        self.parent = [i for i in range(n+1)]
        self.rank = [0 for i in range(n+1)]
    
    def findroot(self,x):
        if x == self.parent[x]:
            return x
        else:
            y = self.parent[x]
            y = self.findroot(self.parent[x])
            return y
    
    def union(self,x,y):
        px = self.findroot(x)
        py = self.findroot(y)
        if px < py:
            self.parent[y] = px
        else:
            self.parent[px] = py
 
    def same_group_or_no(self,x,y):
        return self.findroot(x) == self.findroot(y)
def pow_k(x, n):

    if n == 0:
        return 1

    K = 1
    while n > 1:
        if n % 2 != 0:
            K *= x
        x *= x
        n //= 2

    return K * x
def main():  #startline-------------------------------------------
    n = int(input())
    a = list(map(int, input().split()))
    accum = sum(a)
    ans = 0
    b = [0] * (n + 1)
    for i in range(1, n + 1):
        accum -= a[i]
        b[i] = min(2 * b[i - 1] - a[i], accum)
        if b[i] < 0:
            ans = -1
            break
        ans += b[i] + a[i]
        
    print(ans)
if __name__ == "__main__":
    main() 

Fixed, accepted submission:
import itertools
from collections import deque,defaultdict,Counter
from itertools import accumulate
import bisect
from heapq import heappop,heappush,heapify
import math
from copy import deepcopy
import queue
#import numpy as np

Mod = 1000000007
fact = [1, 1]
factinv = [1, 1]
inv = [0, 1] 
for i in range(2, 10**5 + 1):
    fact.append((fact[-1] * i) % Mod)
    inv.append((-inv[Mod % i] * (Mod // i)) % Mod)
    factinv.append((factinv[-1] * inv[-1]) % Mod)
    
def cmb(n, r, p):
    if (r < 0) or (n < r):
        return 0
    r = min(r, n - r)
    return fact[n] * factinv[r] * factinv[n - r] % p
    
def sieve_of_eratosthenes(n):
    if not isinstance(n,int):
        raise TypeError("n is not int")
    if n<2:
        raise ValueError("n is not effective")
    prime = [1]*(n+1)
    for i in range(2,int(math.sqrt(n))+1):
        if prime[i] == 1:
            for j in range(2*i,n+1):
                if j%i == 0:
                    prime[j] = 0
    res = []
    for i in range(2,n+1):
        if prime[i] == 1:
            res.append(i)
    return res

 
class UnionFind:
    def __init__(self,n):
        self.parent = [i for i in range(n+1)]
        self.rank = [0 for i in range(n+1)]
    
    def findroot(self,x):
        if x == self.parent[x]:
            return x
        else:
            y = self.parent[x]
            y = self.findroot(self.parent[x])
            return y
    
    def union(self,x,y):
        px = self.findroot(x)
        py = self.findroot(y)
        if px < py:
            self.parent[y] = px
        else:
            self.parent[px] = py
 
    def same_group_or_no(self,x,y):
        return self.findroot(x) == self.findroot(y)
def pow_k(x, n):

    if n == 0:
        return 1

    K = 1
    while n > 1:
        if n % 2 != 0:
            K *= x
        x *= x
        n //= 2

    return K * x
def main():  #startline-------------------------------------------
    n = int(input())
    a = list(map(int, input().split()))
    accum = sum(a)
    ans = 1
    b = [0] * (n + 1)
    b[0] = 1
    for i in range(1, n + 1):
        accum -= a[i]
        b[i] = min(2 * b[i - 1] - a[i], accum)
        if b[i] < 0:
            ans = -1
            break
        ans += b[i] + a[i]
    if a[0] == 1:
        ans = -1
    if n == 0:
        if a[0] == 1:
            ans = 1
        else:
            ans = -1
    print(ans)
if __name__ == "__main__":
    main()