#include <bits/stdc++.h>
using namespace std;
const int c = 4005;
long long n, m, t[c], fel[c], dp[c][c], rf[c];
vector<int> sz[c];
void solve(int a, int b, int c) {
  if (a > b) {
    return;
  }
  int pos = a;
  for (int i = a; i <= b; i++) {
    if (t[i] < t[pos]) {
      pos = i;
    }
  }
  sz[c].push_back(pos);
  fel[pos] = c;
  solve(a, pos - 1, pos), solve(pos + 1, b, pos);
}
void dfs(int a) {
  rf[a] = 1;
  for (auto x : sz[a]) {
    dfs(x);
    for (int i = rf[a]; i >= 0; i--) {
      for (int j = rf[x]; j >= 0; j--) {
        dp[a][i + j] = max(dp[a][i + j],
                           dp[a][i] + dp[x][j] + j * (m - j) * (t[x] - t[a]));
      }
    }
    rf[a] += rf[x];
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    cin >> t[i];
  }
  solve(1, n, 0);
  int st = sz[0][0];
  dfs(st);
  cout << dp[st][m] << "\n";
  return 0;
}

### Prompt

Create a solution in Cpp for 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;
const int c = 4005;
long long n, m, t[c], fel[c], dp[c][c], rf[c];
vector<int> sz[c];
void solve(int a, int b, int c) {
  if (a > b) {
    return;
  }
  int pos = a;
  for (int i = a; i <= b; i++) {
    if (t[i] < t[pos]) {
      pos = i;
    }
  }
  sz[c].push_back(pos);
  fel[pos] = c;
  solve(a, pos - 1, pos), solve(pos + 1, b, pos);
}
void dfs(int a) {
  rf[a] = 1;
  for (auto x : sz[a]) {
    dfs(x);
    for (int i = rf[a]; i >= 0; i--) {
      for (int j = rf[x]; j >= 0; j--) {
        dp[a][i + j] = max(dp[a][i + j],
                           dp[a][i] + dp[x][j] + j * (m - j) * (t[x] - t[a]));
      }
    }
    rf[a] += rf[x];
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    cin >> t[i];
  }
  solve(1, n, 0);
  int st = sz[0][0];
  dfs(st);
  cout << dp[st][m] << "\n";
  return 0;
}

```