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;
int N;
long A[1<<17],B[1<<17];
int main()
{
	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;
	}
	else if(A[0]>0)
	{
		cout<<-1<<endl;
		return 0;
	}
	B[N]=A[N];
	for(int i=N;i--;)B[i]=B[i+1]+A[i];
	long ans=0,pre=1;
	for(int i=0;i<=N;i++)
	{
		if(pre>B[i])pre=B[i];
		ans+=pre;
		pre-=A[i];
		if(pre<0)
		{
			cout<<-1<<endl;
			return 0;
		}
		pre*=2;
	}
	cout<<ans<<endl;
}
