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;
  inline int read() {
    int x = 0, flag = 1;
    char c = getchar();
    while (c < '0' || c > '9') {
      if (c == '-') flag = 0;
      c = getchar();
    }
    while (c >= '0' && c <= '9') {
      x = (x << 1) + (x << 3) + c - 48;
      c = getchar();
    }
    return flag ? x : -x;
  }
  const int N = 4e3 + 7;
  struct Node {
    int ls, rs;
  } t[N];
  long long dp[N][N];
  int a[N], top = 0, st[N], sz[N], n, m;
  void dfs(int u) {
    sz[u] = 1;
    if (t[u].ls) {
      dfs(t[u].ls);
      const int v = t[u].ls;
      for (int i = sz[u]; ~i; i--)
        for (int j = 0, rg = min(sz[v], m - i); j <= rg; j++)
          dp[u][i + j] = max(dp[u][i + j], dp[u][i] + dp[v][j] +
                                               1ll * j * (m - j) * (a[v] - a[u]));
      sz[u] += sz[v];
    }
    if (t[u].rs) {
      dfs(t[u].rs);
      const int v = t[u].rs;
      for (int i = sz[u]; ~i; i--)
        for (int j = 0, rg = min(sz[v], m - i); j <= rg; j++)
          dp[u][i + j] = max(dp[u][i + j], dp[u][i] + dp[v][j] +
                                               1ll * j * (m - j) * (a[v] - a[u]));
      sz[u] += sz[v];
    }
  }
  int main() {
    n = read(), m = read();
    for (int i = 1; i <= n; i++) {
      a[i] = read();
      int ret = 0;
      while (top && a[st[top]] >= a[i]) ret = st[top--];
      if (ret) t[i].ls = ret;
      if (top) t[st[top]].rs = i;
      st[++top] = i;
    }
    dfs(st[1]);
    printf("%lld", dp[st[1]][m]);
  }
  