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;
  long long read() {
    long long f = 0, s = 0;
    char c = getchar();
    while (c > '9' || c < '0') f = (c == '-'), c = getchar();
    while (c >= '0' && c <= '9')
      s = (s << 3) + (s << 1) + (c ^ '0'), c = getchar();
    return f ? -s : s;
  }
  const long long N = 4005;
  long long n, m;
  long long a[N];
  long long rt;
  long long c[N][2];
  long long st[N], tp = 0;
  long long f[N][N];
  long long siz[N];
  void dfs(long long u) {
    if (c[u][0]) dfs(c[u][0]);
    if (c[u][1]) dfs(c[u][1]);
    siz[u] = 1;
    if (c[u][0]) {
      long long lw = a[c[u][0]] - a[u];
      for (long long i = siz[u]; i >= 0; i--)
        for (long long j = 0; j <= siz[c[u][0]]; j++)
          f[u][i + j] =
              max(f[u][i + j], f[u][i] + f[c[u][0]][j] + j * (m - j) * lw);
      siz[u] += siz[c[u][0]];
    }
    if (c[u][1]) {
      long long rw = a[c[u][1]] - a[u];
      for (long long i = siz[u]; i >= 0; i--)
        for (long long j = 0; j <= siz[c[u][1]]; j++)
          f[u][i + j] =
              max(f[u][i + j], f[u][i] + f[c[u][1]][j] + j * (m - j) * rw);
      siz[u] += siz[c[u][1]];
    }
  }
  long long ans = 0;
  signed main() {
    n = read();
    m = read();
    for (long long i = 1; i <= n; i++) a[i] = read();
    for (long long i = 1; i <= n; i++) {
      long long now = tp;
      while (now && a[i] < a[st[now]]) now--;
      if (now != tp) c[i][0] = st[now + 1];
      if (now) c[st[now]][1] = i;
      tp = now;
      st[++tp] = i;
    }
    for (long long i = 1; i < tp; i++) c[st[i]][1] = st[i + 1];
    rt = st[1];
    memset(f, 0, sizeof f);
    dfs(rt);
    printf("%lld", f[rt][m]);
    return 0;
  }
  