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;
  mt19937_64 rnd(time(NULL));
  const int N = 5010;
  long long a[N];
  int n, m;
  vector<long long> rec(int l, int r) {
    vector<long long> dp(r - l + 2, 0);
    if (l > r) return dp;
    int mn = l;
    for (int i = l + 1; i <= r; i++) {
      if (a[i] < a[mn]) mn = i;
    }
    auto A = rec(l, mn - 1);
    auto B = rec(mn + 1, r);
    vector<long long> cur(2, 0);
    cur[1] = a[mn] * (m - 1ll);
    for (int i = 0; i < A.size(); i++) {
      for (int j = 0; j < B.size(); j++) {
        for (int t = 0; t < cur.size(); t++) {
          dp[i + j + t] =
              max(dp[i + j + t],
                  A[i] + B[j] + cur[t] - a[mn] * (i * j + t * i + t * j) * 2ll);
        }
      }
    }
    return dp;
  }
  void solve() {
    cin >> n >> m;
    for (int i = 1; i <= n; i++) cin >> a[i];
    auto ans = rec(1, n);
    cout << ans[m] << endl;
  }
  int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    int tt = 1;
    while (tt--) {
      solve();
    }
    return 0;
  }
  