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;

const int nm = 1e5;
int n, a[nm + 1];
long long ans = 0;
long long cap[nm + 1]; //leaf以外のキャパ;
long long node[nm + 2];

int main() {
	cin >> n;
	for (int i = 0; i <= n; i++)
		cin >> a[i];

	cap[0] = 1 - a[0];
	for (int i = 1; i <= n; i++)
		cap[i] = min(cap[i - 1] * 2 - a[i], (long long) 4e18);

	for (int i = 0; i <= n; i++)
		if (cap[i] < 0) {
			cout << -1 << endl;
			return 0;
		}

	for (int i = n; i >= 0; i--) {
		if (node[i + 1] > cap[i] * 2) {
			cout << -1 << endl;
			return 0;
		}
		node[i] = min(node[i + 1], cap[i]) + a[i];
		ans += node[i];
	}
	cout << ans << endl;
}
