/**
 *    author:  tourist
 *    created: 14.07.2019 15:43:48       
**/
#include <bits/stdc++.h>

using namespace std;

vector<long long> a;
int n;

long long Go(int from, int to, long long ca, long long cb) {
  if (to - from == 1) {
    return 0;
  }
  long long best;
  best = (long long) 1e18;
  for (int i = from + 1; i < to; i++) {
    long long cur = a[i] * (ca + cb) + Go(from, i, ca, ca + cb) + Go(i, to, ca + cb, cb);
    best = min(best, cur);
  }
  return best;
}

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  a.resize(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  cout << Go(0, n - 1, 1, 1) + a[0] + a[n - 1] << '\n';
  return 0;
}

### Prompt

Create a solution in CPP for 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
/**
 *    author:  tourist
 *    created: 14.07.2019 15:43:48       
**/
#include <bits/stdc++.h>

using namespace std;

vector<long long> a;
int n;

long long Go(int from, int to, long long ca, long long cb) {
  if (to - from == 1) {
    return 0;
  }
  long long best;
  best = (long long) 1e18;
  for (int i = from + 1; i < to; i++) {
    long long cur = a[i] * (ca + cb) + Go(from, i, ca, ca + cb) + Go(i, to, ca + cb, cb);
    best = min(best, cur);
  }
  return best;
}

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  a.resize(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  cout << Go(0, n - 1, 1, 1) + a[0] + a[n - 1] << '\n';
  return 0;
}

```