description: 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].
solution: #include <bits/stdc++.h>
  using namespace std;
  mt19937 rng(chrono::high_resolution_clock::now().time_since_epoch().count());
  #pragma GCC target("avx2")
  #pragma GCC optimization("O3")
  #pragma GCC optimization("unroll-loops")
  int n, m;
  int a[4007];
  int mn[4007][4007];
  long long dp[4007][4007];
  void f(int l, int r) {
    if (l >= r) {
      return;
    }
    int wh = mn[l][r];
    f(l, wh - 1);
    f(wh + 1, r);
    int tp = 0;
    for (int x = 0; x <= wh - l; ++x) {
      for (int y = 0; y <= r - wh; ++y) {
        long long aux = 0;
        aux += dp[mn[l][wh - 1]][x];
        if (l <= wh - 1) {
          aux += 1LL * (a[mn[l][wh - 1]] - a[mn[l][r]]) * x * (m - x);
        }
        aux += dp[mn[wh + 1][r]][y];
        if (wh + 1 <= r) {
          aux += 1LL * (a[mn[wh + 1][r]] - a[mn[l][r]]) * y * (m - y);
        }
        dp[wh][x + y] = max(dp[wh][x + y], aux);
      }
    }
    for (int i = r - l + 1; i >= 1; --i) {
      dp[wh][i] = max(dp[wh][i], dp[wh][i - 1]);
    }
  }
  void input() {
    cin >> n >> m;
    for (int i = 1; i <= n; ++i) {
      cin >> a[i];
    }
    for (int i = 1; i <= n; ++i) {
      mn[i][i] = i;
      for (int j = i + 1; j <= n; ++j) {
        mn[i][j] = mn[i][j - 1];
        if (a[j] < a[mn[i][j]]) {
          mn[i][j] = j;
        }
      }
    }
  }
  void solve() {
    f(1, n);
    cout << dp[mn[1][n]][m] << "\n";
  }
  int main() {
    ios_base ::sync_with_stdio(false);
    cin.tie(NULL);
    int t;
    t = 1;
    while (t--) {
      input();
      solve();
    }
    return 0;
  }
  