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;
int n,a[100100];
long long sum[100100];
long long ans=0;
int main()
{
	cin>>n;
	for(int i=0;i<=n;i++) cin>>a[i];
	for(int i=n;i>=0;i--) sum[i]=sum[i+1]+a[i];
	long long now=1;ans=0;
	for(int i=0;i<=n;i++)
	{
		if(a[i]>now) return puts("-1"),0;
		ans+=now;now-=a[i];
		now=min(now+now,sum[i+1]);
	}
	cout<<ans<<endl;
}