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 <iostream>
#include <vector>
#include <algorithm>

using namespace std;

int main(void) {
    int64_t N;
    cin >> N;
    vector<int64_t> A(N+2), B(N+2);
    for(int64_t i=0; i<=N; i++) {
        cin >> A[i];
    }
    for(int64_t i=N; i>=0; i--) {
        B[i] = B[i+1] + A[i];
    }

    int64_t ans = 0;
    int64_t now = 1;
    for(int64_t i=0; i<=N; i++) {
        if(A[i] > now) {
            cout << -1 << endl;
            return 0;
        } else {
            ans += now;
            now = (now - A[i]) * 2;
            now = min(now, B[i+1]);
        }
    }

    cout << ans << endl;
    return 0;
}
