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 long long
using namespace std;

const int N = 1e5+10, A = 1e18;

int n, a[N], b[N];

signed main() {
	scanf("%lld", &n);
	b[0] = 1;
	for (int i = 0; i <= n; ++i) {
		scanf("%lld", a+i);
		b[i] -= a[i];
		if (b[i] < 0) {
			puts("-1");
			return 0;
		}
		if (i < n) b[i+1] = min(b[i]*2, A);
	}
	int cur = 0, ans = 0;
	for (int i = n; ~i; --i) {
		cur += a[i];
		ans += cur;
		if (i > 0) cur = min(cur, b[i-1]);
	}
	printf("%lld\n", ans);
}
