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 n, m, f[4001], st[4001], cnt, l[4001], r[4001], ls[4001], rs[4001],
      sz[4001], ans[4001][4001];
  void dfs(long long a, long long c) {
    if (!a) return;
    dfs(l[a], ls[a]);
    dfs(r[a], rs[a]);
    sz[a] = 1ll + sz[l[a]] + sz[r[a]];
    for (long long i = 0ll; i <= sz[l[a]]; i++) {
      if (i > m) break;
      for (long long j = 0ll; j <= sz[r[a]]; j++) {
        if (i + j > m) break;
        ans[a][i + j] = max(ans[a][i + j], ans[l[a]][i] + ans[r[a]][j] +
                                               c * (i + j) * (m - i - j));
        if (i + j + 1ll <= m)
          ans[a][i + j + 1ll] =
              max(ans[a][i + j + 1ll], ans[l[a]][i] + ans[r[a]][j] +
                                           c * (i + j + 1ll) * (m - i - j - 1ll));
      }
    }
  }
  int main() {
    scanf("%lld%lld", &n, &m);
    for (long long i = 1ll; i <= n; i++) {
      scanf("%lld", &f[i]);
      while (cnt) {
        if (f[i] < f[st[cnt]]) {
          l[i] = st[cnt];
          ls[i] = f[st[cnt]] - f[i];
          cnt--;
        } else
          break;
      }
      if (cnt) {
        r[st[cnt]] = i;
        rs[st[cnt]] = f[i] - f[st[cnt]];
      }
      cnt++;
      st[cnt] = i;
    }
    dfs(st[1ll], 0ll);
    printf("%lld\n", ans[st[1ll]][m]);
  }
  