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 long long S = 1 << 22;
  char p0[S], *p1, *p2;
  inline long long read() {
    long long x = 0, f = 0;
    char c =
        (p2 == p1 && (p2 = (p1 = p0) + fread(p0, 1, S, stdin)) == p1 ? EOF
                                                                     : *p1++);
    while (!isdigit(c)) {
      if (c == '-') f = 1;
      c = (p2 == p1 && (p2 = (p1 = p0) + fread(p0, 1, S, stdin)) == p1 ? EOF
                                                                       : *p1++);
    }
    while (isdigit(c))
      x = (x << 1) + (x << 3) + (c & 15),
      c = (p2 == p1 && (p2 = (p1 = p0) + fread(p0, 1, S, stdin)) == p1 ? EOF
                                                                       : *p1++);
    return f ? -x : x;
  }
  const long long MAXN = 4000;
  long long a[MAXN + 10], st[MAXN + 10], top = 0;
  long long lc[MAXN + 10], rc[MAXN + 10];
  long long n, m;
  long long dp[MAXN + 10][MAXN + 10], siz[MAXN + 10];
  void dfs(long long x) {
    if (lc[x]) dfs(lc[x]);
    if (rc[x]) dfs(rc[x]);
    siz[x] = siz[lc[x]] + siz[rc[x]] + 1;
    for (long long i = 0; i <= siz[lc[x]]; ++i) {
      for (long long j = 0; j <= siz[rc[x]]; ++j) {
        dp[x][i + j] =
            max(dp[x][i + j], dp[lc[x]][i] + dp[rc[x]][j] - 2 * a[x] * (i * j));
        dp[x][i + j + 1] = max(
            dp[x][i + j + 1], dp[lc[x]][i] + dp[rc[x]][j] - 2 * a[x] * (i) * (j) +
                                  m * a[x] - 2 * a[x] * (i + j) - a[x]);
      }
    }
  }
  signed main() {
    n = read(), m = read();
    for (long long i = 1; i <= n; ++i) a[i] = read();
    for (long long i = 1; i <= n; ++i) {
      long long k = top;
      while (k && a[st[k]] > a[i]) k--;
      if (k) rc[st[k]] = i;
      if (k < top) lc[i] = st[k + 1];
      st[++k] = i;
      top = k;
    }
    dfs(st[1]);
    long long ans = dp[st[1]][m];
    ;
    cout << ans << endl;
    return 0;
  }
  