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>
#define SIZE 100005
#define INF 1000000000000000LL
using namespace std;
typedef long long int ll;
typedef pair <int,int> P;

ll A[SIZE];
ll lim[SIZE];

int main()
{
	int n;
	scanf("%d",&n);
	for(int i=0;i<=n;i++) scanf("%lld",&A[i]);
	lim[0]=1;
	for(int i=1;i<=n;i++)
	{
		ll m=(lim[i-1]-A[i-1])*2LL;
		if(m<0)
		{
			puts("-1");
			return 0;
		}
		lim[i]=min(INF,m);
	}
	if(lim[n]<A[n])
	{
		puts("-1");
		return 0;
	}
	ll ret=A[n],now=A[n];
	for(int i=n-1;i>=0;i--)
	{
		now+=A[i];
		now=min(now,lim[i]);
		ret+=now;
	}
	printf("%lld\n",ret);
}