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;
typedef long long ll;
typedef double db;
#define Check system("pause")
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)

ll a[100005],b[100005];

int main()
{
	ios::sync_with_stdio(false);
	int n;
	ll ans=0,k,t,w;
	cin>>n;
	for(int i=0;i<=n;i++)
	cin>>a[i];
	for(int i=n;i>=0;i--)
	b[i]=b[i+1]+a[i];
	for(int i=0;i<=n;i++)
	{
		if(i==0) w=1;
		else w<<=1;
		w=min(w,b[i]);
		ans+=w;
		w-=a[i];
		if(w<0)
		{
			cout<<"-1"; return 0;
		}
	}
	cout<<ans;
	return 0;
}


