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++)

using namespace std;
using lint=long long;

int main(){
	int n; scanf("%d",&n);
	vector<int> a(n+1);
	rep(d,n+1) scanf("%d",&a[d]);

	lint sum=0;
	rep(d,n+1) sum+=a[d];

	lint ans=0,f=1;
	rep(d,n+1){
		if(f-a[d]<0){
			return puts("-1"),0;
		}
		ans+=f;
		sum-=a[d];
		f=min(2*(f-a[d]),sum);
	}
	printf("%lld\n",ans);

	return 0;
}
