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

ll a[18];
map<pair<int,int>,ll> mem[18][18];
ll rec(int l, int r, int xl, int xr) {
  if (r-l == 1) return (ll)xl*a[l]+(ll)xr*a[r];
  assert(l != r);
  ll&memo = mem[l][r][{xl,xr}];
  if (!memo) {
    ll ans = 1e18;
    for (int m = l+1; m < r; m++) {
      ans = min(ans, rec(l,m,xl,xl+xr)+rec(m,r,xl+xr,xr)-a[m]*(xl+xr));
    }
    memo = ans+1;
  }
  return memo-1;
}

int main() {
  ios::sync_with_stdio(0); cin.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++)
    cin >> a[i];
  cout << rec(0,n-1,1,1) << endl;
}

### Prompt

Your task is to create 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;

ll a[18];
map<pair<int,int>,ll> mem[18][18];
ll rec(int l, int r, int xl, int xr) {
  if (r-l == 1) return (ll)xl*a[l]+(ll)xr*a[r];
  assert(l != r);
  ll&memo = mem[l][r][{xl,xr}];
  if (!memo) {
    ll ans = 1e18;
    for (int m = l+1; m < r; m++) {
      ans = min(ans, rec(l,m,xl,xl+xr)+rec(m,r,xl+xr,xr)-a[m]*(xl+xr));
    }
    memo = ans+1;
  }
  return memo-1;
}

int main() {
  ios::sync_with_stdio(0); cin.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++)
    cin >> a[i];
  cout << rec(0,n-1,1,1) << endl;
}

```