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 <cstdio>
#include <iostream>
#include <cstring>

typedef long long ll;

ll a[101000], b[101000], ans = 0;

int main()
{
	int n; scanf("%d", &n);
	for(int i = 0; i <= n; i++) scanf("%lld", &a[i]);
	for(int i = n; i >= 0; i--)
	{
		b[i] = (b[i + 1] + 1) / 2 + a[i];
		ans += b[i];
	}
	if(b[0] > 1) return 0 * puts("-1");
	ll w = 0, cnt = 0; int j = 0;
	for(int i = n; i >= 0; i--)
	{
		while(j < i && w + b[i] / 2 > cnt)
		{
			ans += cnt - w;
			j++;
			cnt = cnt * 2 + (b[j] & 1);
		}
		if(j >= i) break;
		w += b[i] / 2;
		ans += (i - j) * (b[i] / 2);
	}
	printf("%lld\n", ans);
}