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>
  #pragma GCC optimize("O3")
  using namespace std;
  const int maxn = 4003;
  int n, m, val[maxn];
  void read() {
    cin >> n >> m;
    for (int i = 1; i <= n; i++) cin >> val[i];
  }
  int lc[maxn], rc[maxn];
  int f(int l, int r) {
    if (l > r) return -1;
    int min_val = INT_MAX, idx = -1;
    for (int i = l; i <= r; i++)
      if (val[i] < min_val) min_val = val[i], idx = i;
    lc[idx] = f(l, idx - 1), rc[idx] = f(idx + 1, r);
    return idx;
  }
  vector<long long> merge(vector<long long> a, vector<long long> b, int ver) {
    vector<long long> ans;
    ans.resize((int)a.size() + (int)b.size());
    for (int i = 0; i < (int)a.size(); i++)
      for (int j = 0; j < (int)b.size(); j++) {
        ans[i + j] = max(ans[i + j], a[i] + b[j] - 2ll * i * j * val[ver]);
        ans[i + j + 1] =
            max(ans[i + j + 1], a[i] + b[j] + (long long)(m - 1) * val[ver] -
                                    2ll * ((long long)i * j + i + j) * val[ver]);
      }
    return ans;
  }
  vector<long long> dfs(int ver) {
    if (ver == -1) return {0};
    return merge(dfs(lc[ver]), dfs(rc[ver]), ver);
  }
  void solve() {
    int root = f(1, n);
    auto ans = dfs(root);
    cout << ans[m] << '\n';
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    read();
    solve();
  }
  