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
#import<bits/stdc++.h>
#define endl '\n'
using namespace std;
typedef long long LL;
LL i,j,k=2,m,n,p,s,a[200040],b[200040];
int main()
{
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	cin>>n;
	for(;i<=n;i++)cin>>a[i];
	if(*a)
	{
		if(n||*a>1)cout<<-1;
		else cout<<1;
		return 0;
	}
	for(i=0;i++<n;)
	{
		k-=a[i];
		if(k<0)
		{
			cout<<-1;
			return 0;
		}
		b[i]=k;
		k*=2;
		if(b[i]>100000000000000)
		{
			i++;
			break;
		}
	}
	p=i;
	for(i=n;i;i--)
	{
		if(i<p)m=min(m,b[i]);
		m+=a[i];
		s+=m;
	}
	cout<<s+1;
}