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>
  const int N = 4086;
  int n, m, rt, tp, st[N], sze[N], ls[N], rs[N];
  long long f[N][N], a[N];
  long long Max(long long a, long long b) { return a > b ? a : b; }
  int Min(int a, int b) { return a < b ? a : b; }
  int Abs(int x) { return x < 0 ? -x : x; }
  long long read() {
    char ch = getchar();
    long long nn = 0, ssss = 1;
    while (ch < '0' || ch > '9') {
      if (ch == '-') ssss *= -1;
      ch = getchar();
    }
    while (ch >= '0' && ch <= '9') {
      nn = nn * 10 + (ch - '0');
      ch = getchar();
    }
    return nn * ssss;
  }
  bool dfs(int np) {
    int lw = Abs(a[ls[np]] - a[np]), rw = Abs(a[rs[np]] - a[np]);
    sze[np] = 1;
    if (ls[np]) {
      dfs(ls[np]);
      for (int i = Min(sze[np], m); i + 1; i--)
        for (int j = Min(sze[ls[np]], m); j + 1; j--)
          f[np][i + j] =
              Max(f[np][i + j], f[np][i] + f[ls[np]][j] + 1ll * j * (m - j) * lw);
      sze[np] += sze[ls[np]];
    }
    if (rs[np]) {
      dfs(rs[np]);
      for (int i = Min(sze[np], m); i + 1; i--)
        for (int j = Min(sze[rs[np]], m); j + 1; j--)
          f[np][i + j] =
              Max(f[np][i + j], f[np][i] + f[rs[np]][j] + 1ll * j * (m - j) * rw);
      sze[np] += sze[rs[np]];
    }
    return true;
  }
  int main() {
    n = (int)read();
    m = (int)read();
    a[rt = n + 1] = 1145141919000;
    for (int i = 1; i <= n; i++) {
      a[i] = read();
      int lt = tp;
      while (tp && a[st[tp]] > a[i]) tp--;
      if (tp) rs[st[tp]] = i;
      if (tp < lt) ls[i] = st[tp + 1];
      st[++tp] = i;
      if (a[i] < a[rt]) rt = i;
    }
    dfs(rt);
    printf("%lld", f[rt][m]);
  }
  