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 <iostream>
using namespace std;

const int mxn = 1e5 + 5;
long long n, a[mxn], sfx[mxn];

int main() {
	cin >> n;
	for (int i=0; i<=n; i++) {
		cin >> a[i];
	}
	for (int i=n; i>=0; i--) {
		sfx[i] = sfx[i + 1] + a[i];
	}
	long long nodes = 1, ans = 0;
	for (int i=0; i<=n; i++) {
		ans += nodes;
		if (a[i] > nodes) {
			cout << -1 << '\n';
			return 0;
		}
		nodes = min(sfx[i + 1], 2 * (nodes - a[i]));
	}
	cout << ans << '\n';
}