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 maxn = 4000 + 10;
  const long long INF = 4e18;
  int m, a[maxn];
  int newInt = 1;
  long long dp[maxn][maxn];
  int sz[maxn];
  int solve(int l, int r) {
    if (l == r) {
      dp[0][0] = 0;
      return 0;
    }
    int v = newInt++;
    dp[v][0] = 0;
    if (l + 1 == r) {
      dp[v][1] = 1LL * (m - 1) * a[l];
      sz[v] = 1;
      return v;
    }
    int idx = l;
    for (int i = l; i < r; i++)
      if (a[i] < a[idx]) idx = i;
    int lc = solve(l, idx);
    int rc = solve(idx + 1, r);
    int sz1 = sz[lc], sz2 = sz[rc];
    for (int i = 1; i <= sz1 + 1; i++)
      dp[v][i] = max(dp[lc][i], dp[lc][i - 1] + 1LL * (m - (2 * i - 1)) * a[idx]);
    for (int i = sz1 + 1; i >= 0; i--)
      for (int j = sz2; j >= 0; j--)
        dp[v][i + j] =
            max(dp[v][i + j], dp[v][i] + dp[rc][j] - 2LL * i * j * a[idx]);
    sz[v] = sz1 + sz2 + 1;
    return v;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    int n;
    cin >> n >> m;
    for (int i = 1; i <= n; i++) cin >> a[i];
    memset(dp, -63, sizeof dp);
    solve(1, n + 1);
    cout << dp[1][m] << '\n';
  }
  