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: Accepted)

#include <bits/stdc++.h>
using namespace std;
#define rep(i,n) for (int i = 0; i < (n); i++)

int main() {
    int N;
    cin >> N;
    
    vector<int> a(N+1);
    rep(i,N+1) {
        cin >> a[i];
    }
    
    vector<long int> d_max(N+1), d_min(N+1), max(N+1);
    bool tf = true;
    
    reverse(a.begin(), a.end());
    
    rep(i,N+1) {
        if(i == 0) {
            d_min[i] = a[0];
            d_max[i] = a[0];
        }
        else {
            d_min[i] = (d_min[i-1] + 1) / 2 + a[i];
            d_max[i] = d_max[i-1] + a[i];
        }
    }
    
    if(d_min[N] > 1)
        tf = false;
    
    reverse(d_max.begin(), d_max.end());
    reverse(a.begin(), a.end());
    
    if(a[N] < 1)
        tf = false;
    
    rep(i,N+1) {
        if(i == 0)
            max[i] = 1;
        else {
            max[i] = min((max[i-1] - a[i-1]) * 2, d_max[i]);
            if(max[i] < max[i-1] - a[i-1])
                tf = false;
        }
    }
    
    long int sum = 0;
    if(tf) {
        rep(i,N+1) {
            sum += max[i];
        }
    }
    else
        sum = -1;
    
    cout << sum << endl;
    
    return 0;
}