#include <bits/stdc++.h>

const int N = 20;

int n, A[N];
std::map<std::pair<int, int>, long long> map[N][N];

long long dfs(int l, int r, int L, int R) {
  if (r == l + 1) return 0;
  if (map[l][r].count({ L, R }))
    return map[l][r][{ L, R }];
  long long ret = 1e18;
  for (int k = l + 1; k < r; ++k)
    ret = std::min(ret, dfs(l, k, L, L + R) + dfs(k, r, L + R, R)
      + 1ll * (L + R) * A[k]);
  return map[l][r][{ L, R }] = ret;
}

int main() {
  std::cin >> n;
  for (int i = 1; i <= n; ++i)
    std::cin >> A[i];
  std::cout << dfs(1, n, 1, 1) + A[1] + A[n] << std::endl;
  return 0;
}

### 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>

const int N = 20;

int n, A[N];
std::map<std::pair<int, int>, long long> map[N][N];

long long dfs(int l, int r, int L, int R) {
  if (r == l + 1) return 0;
  if (map[l][r].count({ L, R }))
    return map[l][r][{ L, R }];
  long long ret = 1e18;
  for (int k = l + 1; k < r; ++k)
    ret = std::min(ret, dfs(l, k, L, L + R) + dfs(k, r, L + R, R)
      + 1ll * (L + R) * A[k]);
  return map[l][r][{ L, R }] = ret;
}

int main() {
  std::cin >> n;
  for (int i = 1; i <= n; ++i)
    std::cin >> A[i];
  std::cout << dfs(1, n, 1, 1) + A[1] + A[n] << std::endl;
  return 0;
}

```