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;

const int N = 100005;
const LL inf = (LL)1e18;;

int n;
LL a[N], ans;

LL dfs(int d, LL v)
{
	v = min(v, inf);
	if (v < a[d]) return -1;
	if (d == n) return ans += a[d], a[d];
	LL w = dfs(d + 1, (v - a[d]) * 2);
	if (w == -1) return -1;
	ans += min(w, v - a[d]) + a[d];
	return a[d] + min(w, v - a[d]);
}

int main()
{
	scanf("%d", &n);
	for (int i = 0; i <= n; i++) scanf("%lld", &a[i]);
	if (dfs(0, 1) == -1) puts("-1");
	else printf("%lld\n", ans);
	return 0;
}