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 Python (3.8.2) (judge verdict: Wrong Answer)

N = int(input())
A = list(map(int, input().split()))
V = [0]*(N+1)
flug = 0
Lsum= sum(A)

V[0] = 1
for i in range(N):
    V[i+1] = (V[i]*2 - A[i+1])
    if V[i+1] <= 0 and i != N-1 :
        flug = 1
    if V[i+1] < 0 and i == N-1 :
        flug = 1
    Lsum = Lsum - A[i+1]
    if V[i+1] > Lsum :
        V[i+1] = Lsum

ans = sum(V)+sum(A)
if A[0] != 0:
    ans = -1
if flug == 1:
   ans = -1
if N == 0:
    ans = -1

print(ans)