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 <iostream>
#include <vector>
#include <algorithm>
#include <math.h>
using namespace std;
typedef long long ll;
const int maxn = 1e5 + 5;
int a[maxn];
int main()
{
	int n;
	ll ans, m, sum = 0;
	cin >> n;
	for (int i = 0; i <= n; i++)
	{
		cin >> a[i];
		sum += a[i];
	}
	if (n == 0 && a[0] != 1 || (n > 0 && a[0] == 1))
	{
		cout<<-1<<endl;
		return 0;
	}
	ans = 1;
	m = 1;
	for (int i = 1; i <= n; i++)
	{
		m = min(m * 2, sum);
		if (a[i] > m)
		{
			cout << -1 << endl;
			return 0;
		}
		ans += m;
		m -= a[i];
		sum -= a[i];
	}
	cout << ans << endl;
	return 0;
}