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>

typedef long long ll;
ll arr[100010], s[100010];

int main() {
	int n;
	scanf("%d", &n);
	for (int i = 0; i <= n; ++i)
		scanf("%lld", arr + i);
	s[n] = arr[n];
	for (int i = n - 1; i >= 0; --i)
		s[i] = s[i + 1] + arr[i];
	ll ans = 0, cur = 1;
if (n == 0) {
puts(arr[0] == 1 ? "1" : "-1");
return 0;
}
	if (arr[0] || !arr[n]) {
		puts("-1");
		return 0;
	}
	for (int i = 0; i < n; ++i) {
		if (cur > s[i + 1])
			cur = s[i + 1];
		if (cur * 2 < arr[i + 1]) {
			puts("-1");
			return 0;
		}
		ans += cur + arr[i];
		cur = cur * 2 - arr[i + 1];	
	}
	printf("%lld", ans + arr[n]);
	return 0;
}