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>

int ri() {
	int n;
	scanf("%d", &n);
	return n;
}

int main() {
	int n = ri();
	int a[n + 1];
	for (auto &i : a) i = ri();
	int64_t sum[n + 1];
	sum[n] = 0;
	for (int i = n; i; i--) sum[i - 1] = sum[i] + a[i];
	int64_t cur = 1;
	int64_t res = 0;
	for (int i = 0; i <= n; i++) {
		res += cur;
		cur -= a[i];
		if (cur < 0) {
			res = -1;
			break;
		}
		cur *= 2;
		cur = std::min(cur, sum[i]);
	}
	printf("%" PRId64 "\n", res);
	return 0;
}
