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;
  const int N = 4e3 + 11;
  int n, m;
  long long a[N];
  vector<pair<int, long long> > e[N];
  void build(int l, int r, int fa) {
    if (l > r) return;
    int pos = l;
    for (int i = l + 1; i <= r; i++)
      if (a[i] < a[pos]) pos = i;
    e[fa].push_back(make_pair(pos, a[pos] - a[fa]));
    build(l, pos - 1, pos);
    build(pos + 1, r, pos);
  }
  int siz[N];
  long long f[N][N], g[N];
  void dfs(int w, int fa) {
    siz[w] = 1;
    for (pair<int, long long> v : e[w]) {
      if (v.first == fa) continue;
      dfs(v.first, w);
      for (int i = 0; i <= siz[w] + siz[v.first]; i++) g[i] = 0;
      for (int i = 0; i <= siz[w]; i++)
        for (int j = 0; j <= siz[v.first]; j++) {
          g[i + j] =
              max(g[i + j], f[w][i] + f[v.first][j] + j * (m - j) * v.second);
        }
      siz[w] += siz[v.first];
      for (int i = 0; i <= siz[w]; i++) f[w][i] = max(f[w][i], g[i]);
    }
  }
  int main() {
    ios::sync_with_stdio(false);
    cin >> n >> m;
    for (int i = 1; i <= n; i++) cin >> a[i];
    build(1, n, 0);
    dfs(e[0][0].first, 0);
    cout << f[e[0][0].first][m] << endl;
    return 0;
  }
  