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

typedef long long ll;
int n;
ll a[19];
ll f(int l, int r, ll xl, ll xr) {
  if (r - l == 1) return a[l] * xl + a[r] * xr;
  ll ans = 1ll << 60;
  for (int i = l + 1; i < r; i++) {
    ans = min(ans, f(l, i, xl, xl + xr) + f(i, r, xl + xr, xr) - (xl + xr) * a[i]);
  }
  return ans;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
  }
  printf("%lld\n", f(1, n, 1, 1));
}

### Prompt

Generate a CPP solution to the following problem:

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;

typedef long long ll;
int n;
ll a[19];
ll f(int l, int r, ll xl, ll xr) {
  if (r - l == 1) return a[l] * xl + a[r] * xr;
  ll ans = 1ll << 60;
  for (int i = l + 1; i < r; i++) {
    ans = min(ans, f(l, i, xl, xl + xr) + f(i, r, xl + xr, xr) - (xl + xr) * a[i]);
  }
  return ans;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
  }
  printf("%lld\n", f(1, n, 1, 1));
}

```