#include <bits/stdc++.h>
using namespace std;

const int N = 20;

typedef long long LL;

int a[N];

LL dfs(int l, int r, int tl, int tr) {
    if (l > r) return 0;
    LL res = 1e18; 
    for (int i = l; i <= r; i++) 
        res = min(res, dfs(l, i - 1, tl, tl + tr) + dfs(i + 1, r, tl + tr, tr) + (LL)a[i] * (tl + tr));
    return res;
}

int main() {
    int n; scanf("%d", &n);
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
    printf("%lld\n", dfs(2, n - 1, 1, 1) + a[1] + a[n]);
    return 0;
}//
### Prompt

Develop a solution in cpp to the problem described below:

There is a stack of N cards, each of which has a non-negative integer written on it. The integer written on the i-th card from the top is A_i.

Snuke will repeat the following operation until two cards remain:

* Choose three consecutive cards from the stack.
* Eat the middle card of the three.
* For each of the other two cards, replace the integer written on it by the sum of that integer and the integer written on the card eaten.
* Return the two cards to the original position in the stack, without swapping them.



Find the minimum possible sum of the integers written on the last two cards remaining.

Constraints

* 2 \leq N \leq 18
* 0 \leq A_i \leq 10^9 (1\leq i\leq N)
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N
A_1 A_2 ... A_N


Output

Print the minimum possible sum of the integers written on the last two cards remaining.

Examples

Input

4
3 1 4 2


Output

16


Input

6
5 2 4 1 6 9


Output

51


Input

10
3 1 4 1 5 9 2 6 5 3


Output

115

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;

const int N = 20;

typedef long long LL;

int a[N];

LL dfs(int l, int r, int tl, int tr) {
    if (l > r) return 0;
    LL res = 1e18; 
    for (int i = l; i <= r; i++) 
        res = min(res, dfs(l, i - 1, tl, tl + tr) + dfs(i + 1, r, tl + tr, tr) + (LL)a[i] * (tl + tr));
    return res;
}

int main() {
    int n; scanf("%d", &n);
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
    printf("%lld\n", dfs(2, n - 1, 1, 1) + a[1] + a[n]);
    return 0;
}//
```