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;
typedef pair<ll,ll> P;
ll a[114514],b[114514];
int main(){
	ll n;
	cin>>n;
	for(int i=0;i<=n;i++)cin>>a[i];
	b[n]=a[n];
	for(int i=n-1;i>=0;i--){
		b[i]=a[i];
		b[i]+=b[i+1];
	}
	ll ans=1,num=1;
	if(a[0]==0){
		for(int i=1;i<=n;i++){
			num=min(2*num,b[i]);
			ans+=num;
			num-=a[i];
			if(num<0||(i!=n&&num==0)){
				cout<<-1;
				return 0;
			}
		}
		cout<<ans;
	}else if(a[0]==1)cout<<(n==0?1:-1);
	else cout<<-1;
	return 0;
}
