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 long ll;
const ll INF=1000000000000000000;

int main(){
	int n;
	int a[100010];
	scanf("%d",&n);
	for(int i=0;i<=n;i++)scanf("%d",&a[i]);
	
	ll CNT=1;
	bool ok=true;
	for(int i=0;i<=n;i++){
		CNT-=a[i];
		if(CNT<0){
			ok=false;
			break;
		}
		CNT*=2;
		CNT=min(CNT,INF);
	}
	if(!ok){
		puts("-1");
		return 0;
	}
	
	ll ret=0;
	ll MAX[100010];
	MAX[0]=1;
	for(int i=1;i<=n;i++){
		MAX[i]=(MAX[i-1]-a[i-1])*2;
		MAX[i]=min(MAX[i],INF);
	}
	ll sum=0;
	for(int i=n;i>=0;i--){
		sum+=a[i];
		ret+=min(MAX[i],sum);
	}
	cout<<ret<<endl;
}