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;
  long long a[4005], v[4005 * 2];
  int m;
  vector<long long> f[4005 * 2];
  int cnt[4005 * 2], tot;
  int son[4005 * 2][2];
  void Max(long long &x, long long y) {
    if (y > x) x = y;
  }
  void solve(int l, int r, int &rt) {
    rt = ++tot;
    f[rt].push_back(0);
    for (int i = l; i <= r; i++) f[rt].push_back(0);
    if (l > r) {
      return;
    }
    if (l == r) {
      v[rt] = a[l];
      cnt[rt] = 1;
      return;
    }
    int mx = a[l], id = l;
    for (int i = l; i <= r; i++) {
      if (a[i] < mx) {
        id = i;
        mx = a[i];
      }
    }
    v[rt] = mx;
    solve(l, id - 1, son[rt][0]);
    solve(id + 1, r, son[rt][1]);
    for (int i = 0; i <= cnt[son[rt][0]]; i++) {
      for (int j = 0; j <= cnt[son[rt][1]]; j++) {
        Max(f[rt][i + j],
            f[son[rt][0]][i] + (v[son[rt][0]] - v[rt]) * (i) * (m - i) +
                f[son[rt][1]][j] + (v[son[rt][1]] - v[rt]) * j * (m - j));
        Max(f[rt][i + j + 1],
            f[son[rt][0]][i] + (v[son[rt][0]] - v[rt]) * (i) * (m - i) +
                f[son[rt][1]][j] + (v[son[rt][1]] - v[rt]) * j * (m - j));
      }
    }
    cnt[rt] = cnt[son[rt][0]] + cnt[son[rt][1]] + 1;
  }
  int main() {
    int n;
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i++) {
      scanf("%lld", &a[i]);
    }
    int rt = 0;
    solve(1, n, rt);
    printf("%lld\n", f[rt][m]);
    return 0;
  }
  