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;
using ll = long long;

const int N = 100005;
const ll oo = 5e17;

int n;
int a[N];
ll r[N];

int main() {
	scanf("%d", &n);
	n++;
	for(int i = 0; i < n; i++) scanf("%d", a + i);
	if(n == 1) return printf("%d\n", a[0] == 1 ? 1 : -1), 0;
	if(a[0]) return printf("-1\n"), 0;
	r[0] = 1;
	for(int i = 1; i < n; i++) {
		r[i] = min(2 * r[i - 1] - a[i], oo);
		if(r[i] < 0) return printf("-1\n"), 0;
	}
	ll ans = 0;
	ll nodes = 0;
	for(int i = n - 1; i >= 0; i--) {
		assert(nodes <= 2 * r[i]);
		nodes = min(nodes, r[i]);
		nodes += a[i];
		ans += nodes;
	}
	printf("%lld\n", ans);
}
