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: Runtime Error)

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
typedef pair<ll,ll> P;

ll N;
ll A[200000];
ll max_n[200000];

int main(){
    cin >> N;
    for(int i=0;i<=N;i++) cin >> A[i];
    max_n[0] = 1;
    for(int i=1;i<=N;i++){
        max_n[i] = (max_n[i-1] - A[i-1]) * 2;
        if(max_n[i] <= 0){
            cout << -1 << endl;
            return 0;
        }
        if(max_n[i] > 1e18){
            max_n[i] = 1e18;
            return -1;
        }
    }
    ll n = 0;
    ll ans = 0;
    for(int i=N;i>=0;i--){
        n += A[i];
        ans += min(max_n[i], n);
    }
    cout << ans << endl;
    return 0;
}