Problem:
Score : 600 points

Problem Statement

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.

Notes

A binary tree is a rooted tree such that each vertex has at most two direct children.

A leaf in a binary tree is a vertex with zero children.

The depth of a vertex v in a binary tree is the distance from the tree's root to v. (The root has the depth of 0.)

The depth of a binary tree is the maximum depth of a vertex in the tree.

Constraints

0 \leq N \leq 10^5

0 \leq A_i \leq 10^{8} (0 \leq i \leq N)

A_N \geq 1

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_0 A_1 A_2 \cdots A_N

Output

Print the answer as an integer.

Sample Input 1

3
0 1 1 2

Sample Output 1

7

Below is the tree with the maximum possible number of vertices. It has seven vertices, so we should print 7.

Sample Input 2

4
0 0 1 0 2

Sample Output 2

10

Sample Input 3

2
0 3 1

Sample Output 3

-1

Sample Input 4

1
1 1

Sample Output 4

-1

Sample Input 5

10
0 0 1 1 2 3 5 8 13 21 34

Sample Output 5

264

Solution in PyPy3 (7.3.0) (judge verdict: Wrong Answer)

import math
N = int(input())
A = list(map(int,input().split()))
Z = A[-1]
A[-1] = 0
flag = 0
B = A[::-1]
if N==0:
    if A[0]!=0:
        print(-1)
    else:
        print(1)
else:
    sita = [0]*(N+1)
    sita[0] = Z
    for i in range(1,N+1):
        X = B[i-1]+sita[i-1]
        if X<=0:
            flag = 1
            break
        if math.log2(X+B[i]) > (N-i):
            sita[i] = 2**(N-1)-B[i]
        else:
            sita[i] = X

    ue = [0]*(N+1)
    ue[0] = 1
    for i in range(1,N+1):
        Y = ue[i-1]*2-A[i]
        if Y>=sita[N-i]:
            ue[i] = sita[N-i]
        else:
            ue[i] = Y
    if flag==1:
        print(-1)
    else:
        print(sum(B)+sum(ue))