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;
#define int long long
const int N=2e5+5;
int a[N],b[N];
signed main(){
	int n,sum=1,x;
	cin>>n;
	for(int i=0;i<=n;++i)
		cin>>a[i];
	for(int i=n;i>=0;--i)
		b[i]=b[i+1]+a[i];
	if(!n) return cout<<(a[0]==1?"1":"-1"),0;
	if(a[0]) return cout<<"-1",0;
	//cout<<1<<endl;
	int res=1;
	for(int i=1;i<=n;++i){
		res=min(res*2,b[i]);
		//cout<<res<<" "<<b[i]<<endl;
		if(res<a[i]) return cout<<"-1",0;
		sum+=res;res-=a[i];
	}
	cout<<sum;
	return 0;
}