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 <string>
#include <vector>
#include <algorithm>
#include <numeric>

using namespace std;
using ll = long long;

int main(){
	ios_base::sync_with_stdio(false);
	int n;
	cin >> n;
	vector<ll> a(n + 1);
	for(int i = 0; i <= n; ++i){ cin >> a[i]; }
	ll answer = 0, cur = 1, target = accumulate(a.begin(), a.end(), 0ll);
	for(int i = 0; i <= n && answer >= 0; ++i){
		if(cur < a[i]){
			answer = -1;
			break;
		}
		answer += cur;
		cur    -= a[i];
		target -= a[i];
		cur     = min(cur * 2, target);
	}
	cout << answer << endl;
	return 0;
}

