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 = 4005;
  int n, m, u, tot = 0;
  int siz[maxn], a[maxn], b[maxn];
  int ch[maxn][2];
  long long f[maxn][maxn];
  inline int build(int l, int r, int sub) {
    if (l > r) return 0;
    int u = ++tot;
    int ptr = -1;
    for (int i = l; i <= r; ++i)
      if (!~ptr || a[i] < a[ptr]) ptr = i;
    siz[u] = 1;
    int p, q;
    p = ch[u][0] = build(l, ptr - 1, a[ptr]);
    siz[u] += siz[p];
    q = ch[u][1] = build(ptr + 1, r, a[ptr]);
    siz[u] += siz[q];
    a[ptr] -= sub, b[u] = a[ptr];
    for (int i = 0; i <= siz[u]; ++i) {
      for (int j = max(0, i - 1 - siz[q]); j <= min(i - 1, siz[p]); ++j) {
        int k = i - 1 - j;
        long long res = f[p][j] + f[q][k];
        f[u][i] = max(f[u][i], res);
      }
      for (int j = max(0, i - siz[q]); j <= min(i, siz[p]); ++j) {
        int k = i - j;
        long long res = f[p][j] + f[q][k];
        f[u][i] = max(f[u][i], res);
      }
      if (u) f[u][i] += 1ll * i * (m - i) * a[ptr];
    }
    return u;
  }
  int main() {
    scanf("%d%d", &n, &m);
    for (int i = 0; i < n; ++i) scanf("%d", a + i);
    int rt = build(0, n - 1, 0);
    printf("%lld\n", f[rt][m]);
    return 0;
  }
  