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<cstdio>
#include<algorithm>
using namespace std;
long long sum[100010];
int a[100010];
int n;
long long ans;
int main(){
	scanf("%d",&n);
	for (int i=0;i<=n;i++) scanf("%d",&a[i]);
	long long s=1;
	sum[n+1]=0;
	for (int i=n;i>=0;i--) sum[i]=sum[i+1]+a[i];
	for (int i=0;i<=n;i++)
	{
		if (s<=sum[i]) ans+=s;
		else ans+=sum[i];
		s-=a[i];
		if (s<0)
		{
			printf("-1\n");
			return 0;
		}
		s=s+s;
		if (s>=1e13) s=1e13;
	}
	 printf("%lld\n",ans);
}