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 ll long long int
#define ff(i,n) for(i=0;i<n;i++)
ll bg=1000000000000000000;
int main()
{ 
       ll n,i,vertex = 0;
		cin>>n;
		ll a[n+1],b[n+1];
		ff(i,n+1)
		   cin>>a[i];
		b[0] = 1-a[0];
		if(b[0]<0)
		{
			cout<<"-1";
			return 0;	
		}   
		for(i=1;i<=n;i++)
		{
			b[i] = min(bg,2*b[i-1]-a[i]);
			if(b[i]<0)
			{
				cout<<"-1";
				return 0;
			}
		}
		b[n] = 0;
		for(i=n-1;i>=0;i--)
			b[i]=min(b[i],a[i+1]+b[i+1]);
		for(i=0;i<=n;i++)
			vertex+=(a[i]+b[i]);
		cout<<vertex;	
	return 0;
}