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;
  const int MOD = 998244353;
  template <class T>
  using vi2 = vector<vector<T>>;
  using ll = long long;
  using pii = pair<int, int>;
  int main() {
    auto solve = [&]() {
      int n, m;
      cin >> n >> m;
      vector<int> a(n + 1), per(n + 1), fa(n + 1), sz(n + 1);
      for (int i = 1, _i = int(n); i <= _i; i++) cin >> a[i];
      function<int(int)> root = [&](int x) {
        return fa[x] ? fa[x] = root(fa[x]) : x;
      };
      auto un = [&](int a, int b) {
        a = root(a), b = root(b);
        fa[a] = b;
      };
      for (int i = 1, _i = int(n); i <= _i; i++) per[i] = i;
      sort(per.begin() + 1, per.end(), [&](int x, int y) { return a[x] > a[y]; });
      vi2<int> g(n + 1);
      for (int i = 1, _i = int(n); i <= _i; i++) {
        int x = per[i];
        if (x != 1 && a[x - 1] > a[x]) {
          int r = root(x - 1);
          g[x].emplace_back(r);
          un(r, x);
        }
        if (x != n && a[x + 1] > a[x]) {
          int r = root(x + 1);
          g[x].emplace_back(r);
          un(r, x);
        }
      }
      int r = per[n];
      vi2<ll> dp(n + 1);
      function<void(int, int)> dfs = [&](int x, int fa) {
        vector<ll> dp1(2, -LLONG_MAX / 3);
        dp1[0] = 0;
        dp1[1] = 1LL * (m - 1) * a[x];
        sz[x] = 1;
        for (int to : g[x]) {
          dfs(to, x);
          vector<ll> tmp1(sz[x] + sz[to] + 1, -LLONG_MAX / 3);
          for (int l = 0, _l = int(sz[x]); l <= _l; l++)
            for (int r = 0, _r = int(sz[to]); r <= _r; r++) {
              tmp1[l + r] =
                  max(dp1[l] + dp[to][r] - 2LL * l * r * a[x], tmp1[l + r]);
            }
          sz[x] += sz[to];
          dp1 = tmp1;
        }
        dp[x] = dp1;
      };
      dfs(r, r);
      cout << dp[r][m] << endl;
    };
    int t = 1;
    while (t--) {
      solve();
    }
    return 0;
  }
  