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;
  template <class T>
  inline T read() {
    T r = 0, f = 0;
    char c = getchar();
    while (!isdigit(c)) f |= c == '-', c = getchar();
    while (isdigit(c)) r = (r << 3) + (r << 1) + (c ^ 48), c = getchar();
    return f ? -r : r;
  }
  const long long N = 4005, INF = 1e9;
  long long n, m, a[N];
  long long sz[N], f[N][N];
  long long stk[N], top, ch[N][2];
  void dfs(long long u) {
    sz[u] = 1;
    for (long long k = 0; k < 2; ++k) {
      long long v = ch[u][k], val = a[v] - a[u];
      if (!v) continue;
      dfs(v);
      for (long long i = min(m, sz[u]); ~i; --i)
        for (long long j = sz[v]; ~j; --j)
          f[u][i + j] = max(f[u][i + j], f[u][i] + f[v][j] + val * j * (m - j));
      sz[u] += sz[v];
    }
  }
  signed main() {
    n = read<long long>(), m = read<long long>();
    for (long long i = 1; i <= n; ++i) {
      a[i] = read<long long>();
      long long k = top;
      while (k && a[stk[k]] > a[i]) --k;
      if (k) ch[stk[k]][1] = i;
      if (k < top) ch[i][0] = stk[k + 1];
      stk[++k] = i, top = k;
    }
    dfs(stk[1]);
    printf("%lld", f[stk[1]][m]);
    return 0;
  }
  