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;
using i64 = int_fast64_t;
#define rep(i, N) for(int i = 0; i < N; i++)

int main(){
    i64 N;
    cin >> N; N++;
    vector<i64> A(N);
    vector<i64> sumA(N);
    rep(i, N) cin >> A[i];
    sumA = A;
    for(int i = N-2; i >= 0; i--) sumA[i] += sumA[i + 1];

    bool can = true;
    i64 B = 1 - A[0];
    if((A[0] > 0 && N != 1) || B < 0) can = false;

    i64 ans = B + sumA[0];
    for(int i = 1; i < N-1; i++){
        i64 b = min(2L*B - A[i], sumA[i+1]);

        if(b < 0) can = false;
        ans += b;
        B = b;
    }
    if(N != 1 && 2L*B < A[N-1]) can = false;

    if(!can) cout << -1 << endl;
    else cout << ans << endl;
}