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 ll long long
using namespace std;
int n;
ll ans;
ll prem[100010];
int a[100010];
int main(){
	cin>>n;
	int x;cin>>x;
	if(n==0){
		if(x==1)cout<<1;
		else cout<<-1;
		return 0;
	}
	if(x!=0){
		cout<<-1;
		return 0;
	}
	for(int i=1;i<=n;i++)cin>>a[i];
	prem[n]=a[n];
	for(int i=n-1;i>=1;i--)
	prem[i]=prem[i+1]+a[i];
	ll la=1;ans=1;
	for(int i=1;i<=n;i++){
		x=a[i];
		if(la*2<x){
			cout<<-1;
			return 0;
		}
		la=la*2-x;
		if(la>prem[i+1])la=prem[i+1];
		if(i==n)la=0;
		ans+=la+a[i];
	}
	cout<<ans;
    return 0;
}






