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;
#define rep(i, n) for(int i=0; i<n; i++)

int main() {
	int N;
	cin >> N;
	vector<int> v(N+1);
	int64_t s = 0;
	rep(i, N+1) {
		cin >> v.at(i);
		s += v.at(i);
	}

	int64_t a=0, b=1, c=0, x, y, z;
	rep(i, N+1) {
		a += b;
		b -= v.at(i);
		s -= v.at(i);
		b = min(b*2, s);
		if ((b <= 0 || s == 0) && i != N) {
			std::cout << "-1" << '\n';
			return 0;
		} else if (b != s && i == N) {
			std::cout << "-1" << '\n';
			return 0;
		}
	}

	std::cout << a << '\n';
}
