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;

int a[100010];
int ll[100010], rr[100010];

signed main () {
	int n; scanf("%lld", &n);
	for (int i = 0; i <= n; i++) {
		scanf("%lld", &a[i]);
	}
	ll[n] = rr[n] = a[n];
	for (int i = n - 1; i >= 0; i--) {
		ll[i] = (ll[i + 1] + 1) / 2 + a[i];
		rr[i] = rr[i + 1] + a[i];
	}
	int l = 1, r = 1;
	long long ans = 0;
	for (int i = 0; i <= n; i++) {
		l = max(l, ll[i]), r = min(r, rr[i]);
		if (l > r || r < a[i]) {
			printf("-1\n");
			return 0;
		}
		ans += r;
		l = l - a[i];
		r = (r - a[i]) * 2;
	}
	printf("%lld\n", ans);
	return 0;
}
