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 N = 4e3 + 1e2 + 7;
  long long n, m, st[N], top, a[N];
  long long f[N][N], tot, g[N];
  long long dfs(long long l, long long r, long long pw) {
    long long p = min_element(a + l, a + r + 1) - a;
    long long x = ++tot;
    for (long long j = 0; j <= r - l + 1; j++) f[x][j] = -1e18;
    f[x][0] = 0;
    f[x][1] = 0;
    long long ls = -1, rs = -1;
    if (l <= p - 1) ls = dfs(l, p - 1, a[l]);
    if (p + 1 <= r) rs = dfs(p + 1, r, a[l]);
    long long sz = 1;
    if (ls != -1) {
      long long szl = p - l;
      long long mnl = *min_element(a + l, a + p);
      for (long long j = 0; j <= szl + sz; j++) g[j] = -1e18;
      for (long long j = 0; j <= szl; j++)
        for (long long k = 0; k <= sz; k++)
          g[j + k] =
              max(g[j + k], f[ls][j] + f[x][k] + (m - j) * j * (mnl - a[p]));
      sz += szl;
      for (long long j = 0; j <= sz; j++) f[x][j] = max(f[x][j], g[j]);
    }
    if (rs != -1) {
      long long szr = r - p;
      long long mnr = *min_element(a + p + 1, a + r + 1);
      for (long long j = 0; j <= szr + sz; j++) g[j] = -1e18;
      for (long long j = 0; j <= szr; j++)
        for (long long k = 0; k <= sz; k++)
          g[j + k] =
              max(g[j + k], f[rs][j] + f[x][k] + (m - j) * j * (mnr - a[p]));
      sz += szr;
      for (long long j = 0; j <= sz; j++) f[x][j] = max(f[x][j], g[j]);
    }
    return x;
  }
  signed main() {
    scanf("%lld%lld", &n, &m);
    for (long long i = 1; i <= n; i++) scanf("%lld", &a[i]);
    dfs(1, n, 0);
    printf("%lld\n", f[1][m]);
  }
  