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>

#define rep(i,n) for(int i=0;i<n;i++)
#define repn(i,n) for(int i=1;i<=n;i++)
#define LL long long
#define pii pair <int,int>
#define fi first
#define se second
#define pb push_back
#define mpr make_pair

using namespace std;

const LL MOD=1e9+7;

LL n,a[100010],cur=1,sum[100010],ans=0;

int main()
{
	//freopen("in.txt","r",stdin);
	cin>>n;
	rep(i,n+1) cin>>a[i];
	for(int i=n;i>=0;i--) sum[i]=sum[i+1]+a[i];
	rep(i,n+1)
	{
		ans+=cur;
		cur-=a[i];
		if(cur<0||cur>sum[i+1])
		{
			puts("-1");
			return 0;
		}
		cur=min(cur*2,sum[i+1]);
	}
	cout<<ans<<endl;
	return 0;
}