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 ll long long
using namespace std;
const int N=1e5+5;
int n,a[N];ll l[N],r[N];
int main()
{
	cin>>n;
	for(int i=0;i<=n;i++)cin>>a[i];
	if(a[n]==0ll){
		cout<<"-1\n";
		return 0;
	}
	l[n]=r[n]=a[n];
	for(int i=n-1;i>=0;i--)
	{
		l[i]=(l[i+1]+1ll)/2;
		r[i]=r[i+1];
		l[i]+=a[i];
		r[i]+=a[i];
	}
	if(l[0]>1ll){
		cout<<"-1\n";
		return 0;
	}
	ll ans=1;
	r[0]=1;
	for(int i=1;i<=n;i++)
	{
		l[i]=max(l[i],l[i-1]-a[i-1]);
		r[i]=min(r[i],2ll*(r[i-1]-a[i-1]));
		if(r[i]<l[i])
		{
			cout<<"-1\n";
			return 0;
		}
		ans+=r[i];
	}
	cout<<ans<<"\n";
	return 0;
}