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<iostream>
using namespace std;

typedef long long ll;
ll a[100005];
ll inf[100005];
ll inff=1e18+7;

int main()
{
	int n;
	cin>>n;
	for(int i=0;i<=n;i++){
		cin>>a[i];
	}
	if(n==0){
		if(a[0]==1){
			cout<<"1"<<endl;
		}
		else {
			cout<<"-1"<<endl;
		}
		return 0;
	}
	inf[0]=1;
	for(int i=1;i<=n;i++){
		inf[i]=max(0ll,min(inff,(inf[i-1]-a[i-1])*2));
	}
	ll sum=0;
	ll temp=0;
	for(int i=n;i>=0;i--){
		if((temp+1)/2+a[i]>inf[i]){
			cout<<"-1"<<endl;
			return 0;
		}
		temp=min(inf[i],temp+a[i]);
		sum+=temp;
	}
	cout<<sum<<endl;
	return 0;
}