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 ll long long
const ll MOD = 1000000007;

int main() {
	ll N;
	cin >> N;
	vector<ll> A(N + 1), R(N + 1);
	for (int i = 0;i <= N;i++) {
		cin >> A[i];
	}
	R[N] = A[N];
	for (int i = N - 1;i >= 0;i--) {
		R[i] = R[i + 1] + A[i];
	}
	ll le = 1, ans = 0;
	for (int i = 0;i <= N;i++) {
		if (le < A[i]) {
			cout << -1 << endl;
			return 0;
		}
		if (le > R[i]) {
			le = R[i];
		}
		ans += le;
		le -= A[i];
		le *= 2;
	}
	cout << ans << endl;
}