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;



long long int inf = 1e17;

long long int a[100010];
long long int can[100010],tot[100010];
int main(){
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	int n;
	cin>>n;
	bool ok=true;
	for(int i=0;i<=n;i++){
		cin>>a[i];
		if(i==0){
			if(a[i]>1)	ok=false;
			can[i]=1-a[i];
		}
		else{
			can[i]=2*can[i-1]-a[i];
			if(can[i]<0)	ok=false;
			if(can[i]>inf)	can[i]=inf;
		}
	}
	if(!ok)	return cout<<-1<<endl,0;
	long long int ans=0;
	for(int i=n;i>=0;i--){
		if(i==n){
			tot[i]=a[i];
		}
		else{
			tot[i]=min(tot[i+1],can[i])+a[i];
		}
		ans+=tot[i];
	}
	cout<<ans<<endl;
}

