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<cstdio>
#include<algorithm>
using namespace std;
typedef long long ll;
ll n,l[100005],r[100005],a[100005],ans=0,x;
int main()
{
	scanf("%lld",&n);
	for(int i=0;i<=n;i++)scanf("%lld",&a[i]);
	ans=0;
	if(n==0)
	{
		if(a[0]==1)printf("1\n");
		else printf("-1\n");
		return 0;
	}
	l[n]=r[n]=a[n];
	for(int i=n-1;i>=0;i--)
	{
		l[i]=a[i]+(l[i+1]+1)/2;
		r[i]=a[i]+r[i+1];
	}
	if(l[0]>1||r[0]<1)
	{
		printf("-1\n");
		return 0;
	}
	x=1;
	for(int i=0;i<=n;i++)
	{
		ans+=x;
		x-=a[i];
		x=min(2ll*x,r[i+1]);
	}
	printf("%lld\n",ans);
	return 0;
}