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 C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

int main(){
    long long N;
    cin >> N;
    vector<long long> A(N+1);
    long long sum = 0;
    for(int i = 0; i < N+1;i++){
        cin >> A[i];
        sum += A[i];
    }
    long long k =1;
    long long check = 0;
    for(int i = 0; i < N+1;i++){
        if(k < A[i]){
            check = -1;
            cout << -1 << endl;
            break;
        }
        k = 2*(k - A[i]);
    }
    //cout << sum << endl;
    long long node_sum = 1;
    long long node = 1;
    if(check != -1){
        sum -= A[0];
            if(N > 0){
            for(int i = 1;i < N+1;i++){
                node_sum *= 2;
                if(node_sum > sum){
                    node_sum = sum;
                }
                sum -= A[i];
                node += node_sum;
                node_sum -= A[i];
            }
            cout << node << endl;
            }
    }
}