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;
typedef long long ll;
int main() {
	int N;
	cin >> N;
	vector<ll> A(N+1), B(N+1);
	for (auto& x : A) cin >> x;
	B[N] = A[N];
	for (int i = N-1; i >= 0; -- i) B[i] = B[i+1] + A[i];
	for (int i = 0; i < 50 && i <= N; ++ i) {
		B[i] = min(B[i], 1LL<<i);
	}
	ll hoge = 1;
	ll res = 0;
	for (int i = 0; i <= N; ++ i) {
		hoge = min(hoge, B[i]);
		res += hoge;
		if (A[i] > hoge) {
			cout << -1 << endl;
			return 0;
		}
		hoge -= A[i];
		hoge = hoge*2;
	}
	cout << res << endl;
}
