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;
int n,a[111111];
long long ans,l[111111],r[111111],cur,x,y;
int main()
{
	scanf("%d",&n);
	for (int i=0;i<=n;i++) scanf("%d",&a[i]);
	l[n]=r[n]=a[n];
	for (int i=n-1;i>=0;i--)
	{
		l[i]=(l[i+1]+1)/2+a[i];
		r[i]=r[i+1]+a[i];
	}
	if (l[0]<=1 && r[0]>=1);
	else 
	{
		printf("-1\n");
		return 0;
	}
	ans=cur=1;
	for (int i=1;i<=n;i++)
	{
		x=cur;y=cur*2;
		y=min(y,r[i]);ans+=y;
		cur=y-a[i];
	}
	printf("%lld\n",ans);
	return 0;
}