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>
#define maxn 100010
#define LL long long
using namespace std;
int a[maxn];
LL sum[maxn];
int main(){
	int i,n;
	LL num=1,ans=0;
	scanf("%d",&n);
	for(i=0;i<=n;i++)scanf("%d",&a[i]);
	for(i=n;i>=0;i--)sum[i]=sum[i+1]+a[i];
	for(i=0;i<=n;i++){
		if(num<a[i]){printf("-1\n");return 0;}
		ans+=num;
		num-=a[i];
		num=min(num<<1,sum[i+1]);
	}
	printf("%lld\n",ans);
	return 0;
}