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
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }

signed main(){
	int n;
	cin >> n;

	int a[100010];
	for(int i = 0;i < n+1;i++) cin >> a[i];

	int sum = 0;
	for(int i = 0;i < n+1;i++){
		sum += a[i];
	}

	int now = 1;

	int ans = 0;
	for(int i = 0;i < n+1;i++){
		ans += now;
		now -= a[i];
		sum -= a[i]; 
		if(now < 0){
			cout << -1 << endl;
			return 0;
		}
		now = min(now*2, sum);
	}

	cout << ans << endl;

	return 0;
}
