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 <stdio.h>

int x[100010];
long long int s[100010];
int main()
{
	int a;
	scanf("%d",&a);
	for(int i=0;i<=a;i++) scanf("%d",&x[i]);
	for(int i=a;i>=0;i--) s[i] = s[i+1] + x[i];
	
	long long int S = 1;
	long long int ans = 0;
	for(int i=0;i<=a;i++)
	{
		ans += S;
		if(S<x[i])
		{
			printf("-1");
			return 0;
		}
		S -= x[i];
		S*=2;
		S = S<s[i+1]?S:s[i+1];
	}
	printf("%lld",ans);
}
