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>
int n;
long long a[100001];
long long r[100001];
long long b[100001];
long long x, ans;
int main() {
	scanf("%d", &n);
	for (int i = 0; i <= n; i++) {
		scanf("%lld", a + i);
	}
	r[n] = a[n];
	for (int i = n - 1; i >= 0; i--) {
		r[i] = r[i + 1] + a[i];
	}
	if (n == 0) {
		if (a[0] == 1)printf("1\n");
		else printf("-1\n");
		return 0;
	}
	if (a[0] != 0) {
		printf("-1\n");
		return 0;
	}
	b[0] = 1;
	for (int i = 1; i <= n; i++) {
		x = b[i - 1] * 2;
		if (x < a[i]) {
			printf("-1\n");
			return 0;
		}
		if (r[i] < x)x = r[i];
		b[i] = x - a[i];
		ans += b[i] + a[i];
	}
	printf("%lld\n", ans + 1);
}