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>

using namespace std;

long long n, a[100005], b[100005], c[100005], cur, ans;

int main() {
	cin >> n;
	for (int i = 0; i <= n; i++) {
		cin >> a[i];
	}
	if (a[0] != 0 && n != 0 || n == 0 && a[0] >= 2) {
		cout << -1;
		return 0;
	}
	cur = 1;
	b[0] = 1;
	for (int i = 1; i <= n; i++) {
		cur = cur * 2 - a[i];
		if (cur < 0) {
			cout << -1;
			return 0;
		}
		if (cur > 1e18) cur = 1e18;
		b[i] = cur + a[i];
	}
	c[n] = a[n];
	cur = a[n];
	for (long long i = n - 1; i >= 0; i--) {
		cur += a[i];
		c[i] = cur;
	}
	for (int i = 0; i <= n; i++) {
		ans += min(b[i], c[i]);
	}
	cout << ans;
	return 0;
}