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 <stack>
#include <iostream>
#include <cstdio>
#include <vector>
#include <algorithm>
#include <cmath>
#include <array>
#include <iomanip>
#include <string>
using namespace std;
int gcd(int a, int b)
{
   if (a%b == 0)
   {
       return(b);
   }
   else
   {
       return(gcd(b, a%b));
   }
}
int mod = 1000000000 + 7;

int main(){
    //std::stack<int> stack;
    int n;
    cin >> n;
    vector<long> a(n+1);
    vector<long> b(n+1);
    vector<long> um(n+1);
    for(int i = 0; i <= n; i++){
        cin >> a[i];
    }
    um[n] = a[n];
    for(int i = n-1; i >= 0; i--){
        um[i] = um[i+1] + a[i];
    }
    long long ans = 0;
    b[0] = 1 - a[0];
    bool jadge = true;
    for(int i = 0; i < n; i++){
        if(a[i+1] > 2 * b[i]){
            jadge = false;
            break;
        }else{
            b[i+1] = 2*b[i] - a[i+1];
            b[i+1] = min(b[i+1], um[i+2]);
        }
    }
    if(jadge){
        for(int i = 0; i < n; i++){
            ans += (a[i]+b[i]);
        }
        ans += a[n];
    }
    
    cout << ans << endl;
     return 0;
}