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;

void solve() {
	int n;
	cin >> n;
	vector<int> v(n + 1);
	for (int i = 0; i <= n; i++) {
		cin >> v[i];
	}
	vector<long long> lim(n + 1);
	long long sum = 0;
	for (int i = n; i >= 0; i--) {
		sum += v[i];
		lim[i] = sum;
	}
	long long vert = 1, bot = 1, dep = 0;
	for (int i = 0; i <= n; i++) {
		int x = v[i];
		if (x <= bot) {
			bot -= x;
			bot = min(lim[++dep], bot * 2);
			vert += bot;
		} else {
			cout << "-1\n";
			return;
		}
	}
	cout << vert << "\n";
}

int main() {
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	solve();
	return 0;
}