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;
long long int a[1000100];
long long int b[1000100];
int main()
{
	long long int n,i,ans = 0;
	memset(a,0,sizeof(a));
	memset(b,0,sizeof(b));
	scanf("%lld",&n);
	for(i=0;i<=n;i++)
	{
		scanf("%lld",&a[i]);
		b[i]=1e18;
	}
	b[0]=1;
	if(b[0]<a[0])
	{
		printf("-1");
		return 0;
	}
	for(i=0;i<n;i++)
	{
		b[i+1]= (b[i]-a[i])*2;
		if(b[i]<0)
		{
			printf("-1");
			return 0;
		}
		if(b[i]>=1e18) break;
	}
	if(a[n]>b[n])
	{
		printf("-1");
		return 0;
	}
	long long int now;
	b[n]=ans=now=a[n];

	for(i=n-1;i>=0;i--)
	{
		now=min(now,b[i]-a[i])+a[i];
		ans+=now;
	}
	printf("%lld\n",ans);
}