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;

const int MAXN = 100005;

int n, a[MAXN];
long long R[MAXN];

int main(){
	scanf("%d", &n);
	
	for(int i = 0; i <= n; i++){
		scanf("%d", &a[i]);
		
		if(i == 0){
			R[i] = 1;
		}else{
			R[i] = (R[i - 1] - a[i - 1]) * 2;
		}
		R[i] = min(R[i], 1LL << 60);
		
		if(R[i] < a[i]){
			printf("-1\n");
			return 0;
		}
	}
	
	long long res = 0, c = a[n];
	for(int i = n; i >= 0; i--){
		res += c;
		
		if(i > 0){
			c = min(c + a[i - 1], R[i - 1]);
		}
	}
	printf("%lld\n", res);
		
	return 0;
}
