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;
  void debug() { cout << endl; }
  template <class T, class... U>
  void debug(T a, U... b) {
    cout << a << " ";
    debug(b...);
  }
  template <class T>
  void pary(T l, T r) {
    while (l != r) {
      cout << *l << " ";
      l++;
    }
    cout << endl;
  }
  const long long inf = (1LL << 60);
  vector<pair<int, int> > adj[4005];
  int a[4005], par[4005];
  int m;
  vector<long long> dfs(int n, int pa) {
    vector<long long> ret(2, 0);
    for (auto v : adj[n]) {
      if (v.first != pa) {
        vector<long long> tmp = dfs(v.first, n);
        int siz = ret.size() + tmp.size() - 1;
        vector<long long> cur(siz, -inf);
        for (int i = 0; i < siz; i++) {
          long long val = -inf;
          for (int j = max(i - (int)ret.size() + 1, 0);
               j <= min(i, (int)tmp.size() - 1); j++) {
            val =
                max(val, tmp[j] + (long long)v.second * j * (m - j) + ret[i - j]);
          }
          cur[i] = val;
        }
        ret = cur;
      }
    }
    return ret;
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    int n;
    cin >> n >> m;
    for (int i = 1; i <= n; i++) cin >> a[i];
    stack<int> stk;
    for (int i = 1; i <= n; i++) {
      int prv = 0;
      while (stk.size() && a[i] <= a[stk.top()]) {
        prv = stk.top();
        stk.pop();
      }
      if (prv) par[prv] = i;
      if (stk.size()) par[i] = stk.top();
      stk.push(i);
    }
    for (int i = 1; i <= n; i++) {
      if (par[i]) {
        adj[i].push_back({par[i], a[i] - a[par[i]]});
        adj[par[i]].push_back({i, a[i] - a[par[i]]});
      }
    }
    vector<long long> ans = dfs(1, 0);
    cout << ans[m] << "\n";
  }
  