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 int unsigned long long
using namespace std;

int n,tot=0;
int a[100005],b[100005],c[100005];

int p[100005][205];

signed main()
{
	cin>>n;
	n++;
	for (int i=1;i<=n;i++)  cin>>a[i];
	
	b[1]=1,c[n]=a[n];
	for (int i=2;i<=n;i++)  b[i]=(b[i-1]-a[i-1])*2;
	for (int i=n-1;i>=1;i--)  c[i]=c[i+1]+a[i];
	for (int i=1;i<=n;i++)  tot+=min(b[i],c[i]);
	for (int i=1;i<=n;i++)
	{
		if (a[i]>b[i])  return cout<<-1<<endl,0;
	}
	cout<<tot<<endl;
	
	return 0;
}