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>

using namespace std;

long long solve(const vector<int>& A){
    long long res = 1;
    long long rest = 1 - A[0];
    if(rest < 0) return -1;
    vector<long long> req(A.size(), 0);
    req.back() = A.back();
    for(int i=A.size()-2;i>=0;i--) req[i] = req[i+1] + A[i];
    for(int i=1;i<A.size();i++){
        rest = min(req[i], 2*rest);
        if(rest < A[i]) return -1;
        res += rest;
        rest -= A[i];
    }
    return res;
}

int main(){
    int N; cin >> N;
    vector<int> A(N+1);
    for(auto& a : A) cin >> a;
    cout << solve(A) << endl;
}