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 <iostream>
#include <vector>

using namespace std;

int main()
{
    size_t N;
    vector<size_t> a;

    cin >> N;
    a.resize(N+1);
    for(size_t i = 0; i <= N; ++i)
    {
        cin >> a[i];    
    }

    vector<size_t> b(N+1, 0);

    // 0 での初期化
    b[0] = 1 - a[0];    

    for(size_t i = 1; i <= N; ++i)
    {
        if( a[i] > 2*b[i-1])
        {
            cout << "-1" << endl;
            return 0;    
        }

        size_t s = 2 * b[i-1] - a[i];
        size_t t = 0;
        for(size_t j = i+1; j <= N; ++j)    
        {
            t += a[j];
        }
        
        b[i] =  (s < t) ? s : t;
        if(b[i-1] > a[i] + b[i])
        {
            cout << "-1" << endl;
            return 0;    
        }
    }

    size_t total = 0;
    for(size_t i = 0; i <= N; ++i)
    {
        total += b[i] + a[i];
    }

    cout << total << endl;
    return 0;    
    
}