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;
typedef long double ld;
const int N=1000005;
int n,i,a[N];
long long s,ans,b[N];
int main()
{
	scanf("%d",&n);
	for(i=0;i<=n;++i)
		scanf("%d",&a[i]);
	b[0]=1-a[0];
	if(b[0]<0)
	{
		puts("-1");
		return 0;
	}
	for(i=1;i<=n;++i)
	{
		b[i]=b[i-1]*2-a[i];
		b[i]=min(b[i],10000000000000000ll);
		if(b[i]<0)
		{
			puts("-1");
			return 0;
		}
	}
	for(i=n;i>=0;--i)
	{
		ans+=min(1ll*b[i],s)+a[i];
		s+=a[i];
	}
	cout<<ans;
}
