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;
typedef double DB;

const int maxn = (int)1e5 + 3;
int n, a[maxn];

LL solve() {
	static LL b[maxn];
	b[n] = 0;
	for(int i = n; i > 0; --i)
		b[i - 1] = b[i] + a[i];
	LL ret = 0, cur = 1;
	for(int i = 0; i <= n; ++i) {
		if(cur < a[i])
			return -1;
		ret += cur;
		cur -= a[i];
		if(cur > b[i])
			return -1;
		cur = min(cur << 1, b[i]);
	}
	return !cur ? ret : -1;
}

int main() {
	scanf("%d", &n);
	for(int i = 0; i <= n; ++i)
		scanf("%d", a + i);
	printf("%lld\n", solve());
	return 0;
}