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;
  using ll = long long;
  const int MAX_N = 4000;
  int a[1 + MAX_N];
  vector<int> gr[1 + MAX_N];
  int create_tree(int lb, int rb) {
    if (lb == rb) return lb;
    int id = lb;
    for (int i = lb + 1; i <= rb; i++)
      if (a[id] > a[i]) id = i;
    if (lb < id) gr[id].push_back(create_tree(lb, id - 1));
    if (rb > id) gr[id].push_back(create_tree(id + 1, rb));
    return id;
  }
  const ll INF = 1e18;
  ll dp[1 + MAX_N][1 + MAX_N];
  int sz[1 + MAX_N];
  void upd(ll &a, ll b) {
    if (a < b) a = b;
  }
  int m;
  void calc_dp(int node) {
    dp[node][0] = 0;
    dp[node][1] = 1ll * (m - 1) * a[node];
    sz[node] = 1;
    for (int son : gr[node]) {
      calc_dp(son);
      for (int sz_node = sz[node]; sz_node >= 0; sz_node--)
        for (int sz_son = sz[son]; sz_son >= 0; sz_son--)
          upd(dp[node][sz_node + sz_son], dp[node][sz_node] + dp[son][sz_son] -
                                              2ll * sz_son * sz_node * a[node]);
      sz[node] += sz[son];
      if (sz[node] > m) sz[node] = m;
    }
  }
  void solve_test() {
    int n;
    cin >> n >> m;
    for (int i = 1; i <= n; i++) cin >> a[i];
    int root = create_tree(1, n);
    for (int i = 1; i <= n; i++)
      for (int j = 0; j <= m; j++) dp[i][j] = -INF;
    calc_dp(root);
    cout << dp[root][m] << "\n";
  }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int t = 1;
    while (t--) solve_test();
    return 0;
  }
  