#include <bits/stdc++.h>
using namespace std;
long long int a[4005];
long long int dp[4005][4005];
int m;
int dfs(int l, int r) {
  if (l > r) return 0;
  long long int mi = 1e18;
  long long int root = -1;
  for (int i = l; i <= r; i++) {
    if (a[i] < mi) {
      mi = a[i];
      root = i;
    }
  }
  int root_l = dfs(l, root - 1);
  int root_r = dfs(root + 1, r);
  int sz = r - l + 1;
  for (int i = 0; i <= sz; i++) {
    int sz_l = root - l;
    int sz_r = r - root;
    dp[root][i] = -1e18;
    for (int j = max(0, i - sz_r - 1); j <= min(sz_l, i); j++) {
      dp[root][i] = max(dp[root][i], dp[root_l][j] + dp[root_r][i - j] -
                                         2ll * j * (i - j) * a[root]);
      if (i - 1 - j >= 0)
        dp[root][i] = max(dp[root][i], dp[root_l][j] + dp[root_r][i - 1 - j] -
                                           2ll * (j + 1) * (i - j) * a[root] +
                                           a[root] + 1ll * m * a[root]);
    }
  }
  return root;
}
int main(void) {
  int n;
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%lld", &a[i]);
  for (int i = 0; i <= n; i++) {
    dp[0][i] = -1e18;
  }
  dp[0][0] = 0;
  int root = dfs(1, n);
  printf("%lld\n", dp[root][m]);
  return 0;
}

### Prompt

Your task is to create a cpp solution to the following problem:

Alice has an integer sequence a of length n and all elements are different. She will choose a subsequence of a of length m, and defines the value of a subsequence a_{b_1},a_{b_2},…,a_{b_m} as $$$∑_{i = 1}^m (m ⋅ a_{b_i}) - ∑_{i = 1}^m ∑_{j = 1}^m f(min(b_i, b_j), max(b_i, b_j)), where f(i, j) denotes \min(a_i, a_{i + 1}, \ldots, a_j)$$$.

Alice wants you to help her to maximize the value of the subsequence she choose.

A sequence s is a subsequence of a sequence t if s can be obtained from t by deletion of several (possibly, zero or all) elements.

Input

The first line contains two integers n and m (1 ≤ m ≤ n ≤ 4000).

The second line contains n distinct integers a_1, a_2, …, a_n (1 ≤ a_i < 2^{31}).

Output

Print the maximal value Alice can get.

Examples

Input


6 4
15 2 18 12 13 4


Output


100


Input


11 5
9 3 7 1 8 12 10 20 15 18 5


Output


176


Input


1 1
114514


Output


0


Input


2 1
666 888


Output


0

Note

In the first example, Alice can choose the subsequence [15, 2, 18, 13], which has the value 4 ⋅ (15 + 2 + 18 + 13) - (15 + 2 + 2 + 2) - (2 + 2 + 2 + 2) - (2 + 2 + 18 + 12) - (2 + 2 + 12 + 13) = 100. In the second example, there are a variety of subsequences with value 176, and one of them is [9, 7, 12, 20, 18].

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long int a[4005];
long long int dp[4005][4005];
int m;
int dfs(int l, int r) {
  if (l > r) return 0;
  long long int mi = 1e18;
  long long int root = -1;
  for (int i = l; i <= r; i++) {
    if (a[i] < mi) {
      mi = a[i];
      root = i;
    }
  }
  int root_l = dfs(l, root - 1);
  int root_r = dfs(root + 1, r);
  int sz = r - l + 1;
  for (int i = 0; i <= sz; i++) {
    int sz_l = root - l;
    int sz_r = r - root;
    dp[root][i] = -1e18;
    for (int j = max(0, i - sz_r - 1); j <= min(sz_l, i); j++) {
      dp[root][i] = max(dp[root][i], dp[root_l][j] + dp[root_r][i - j] -
                                         2ll * j * (i - j) * a[root]);
      if (i - 1 - j >= 0)
        dp[root][i] = max(dp[root][i], dp[root_l][j] + dp[root_r][i - 1 - j] -
                                           2ll * (j + 1) * (i - j) * a[root] +
                                           a[root] + 1ll * m * a[root]);
    }
  }
  return root;
}
int main(void) {
  int n;
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%lld", &a[i]);
  for (int i = 0; i <= n; i++) {
    dp[0][i] = -1e18;
  }
  dp[0][0] = 0;
  int root = dfs(1, n);
  printf("%lld\n", dp[root][m]);
  return 0;
}

```