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<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int N=100005;
int n;
int a[N];
long long f[N],g[N];

int main()
{
	scanf("%d",&n);
	for(int i=0;i<=n;i++)
		scanf("%d",&a[i]);
	if(a[0]>1)
	{
		printf("-1");
		return 0;
	}
	g[n]=a[n];
	for(int i=n-1;i>=0;i--)
		g[i]=g[i+1]+a[i];
	f[0]=1;
	for(int i=1;i<=n;i++)
	{
		f[i]=min((f[i-1]-a[i-1])*2,g[i]);
		if(a[i]>f[i])
		{
			printf("-1");
			return 0;
		}
	}
	long long ans=0;
	for(int i=0;i<=n;i++)
		ans+=f[i];
	printf("%lld",ans);
	return 0;
}