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<stdio.h>
#include<vector>
#include<algorithm>
using namespace std;
typedef long long ll;
ll a[202020], r[202020];
int main()
{
	int num;
	scanf("%d", &num);
	for (int i = 0; i <= num; i++)scanf("%lld", &a[i]);
	for (int j = num; j >= 0; j--)r[j] = r[j + 1] + a[j];
	ll ans = 0;
	ll now = 1;
	for (int i = 0; i <= num; i++)
	{
		ans += now;
		if (now < a[i])
		{
			printf("-1\n");
			return 0;
		}
		now -= a[i];
		now = min(now * 2, r[i + 1]);
	}
	printf("%lld\n", ans);
}