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.

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.

Examples

Input

3
0 1 1 2


Output

7


Input

4
0 0 1 0 2


Output

10


Input

2
0 3 1


Output

-1


Input

1
1 1


Output

-1


Input

10
0 0 1 1 2 3 5 8 13 21 34


Output

264
#include <bits/stdc++.h>
using namespace std;
#define int long long
signed main() {
	int N;
    cin >> N;
    vector<int>A(1+N);
    vector<int>B(1+N);
    for(int i = 0; i <= N; i++) {
        cin >> A[i];
    }
    B[N] = 0;
    for(int i = N; i > 0; i--) {
        B[i-1] = B[i]+A[i];
    }
    int ans = 0;
    int cnt = 1;
    for(int i = 0; i <= N; i++) {
        ans+=cnt;
        cnt-=A[i];
        if(cnt < 0) {
            ans = -1;
            break;
        }
        cnt*=2;
        cnt = min(cnt,B[i]);
    }
    cout << ans << endl;
}