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 MAXN = 4010, INF = 1e18;
  int n, m, a[MAXN];
  int st[MAXN], top;
  int root, lc[MAXN], rc[MAXN];
  int sz[MAXN];
  long long f[MAXN][MAXN], ans;
  void build() {
    for (int i = (1); i <= (n); i++) {
      int j = 0;
      while (top && a[st[top]] > a[i]) {
        j = st[top];
        top--;
      }
      if (!top)
        root = i;
      else
        rc[st[top]] = i;
      lc[i] = j;
      st[++top] = i;
    }
  }
  void dfs1(int u) {
    if (!u) return;
    sz[u] = 1;
    dfs1(lc[u]);
    dfs1(rc[u]);
    sz[u] += sz[lc[u]];
    sz[u] += sz[rc[u]];
  }
  void dfs2(int u) {
    if (!u) return;
    dfs2(lc[u]);
    dfs2(rc[u]);
    for (int i = (0); i <= (min(m, sz[lc[u]])); i++) {
      for (int j = (0); j <= (min(m - i, sz[rc[u]])); j++) {
        f[u][i + j] = max(
            f[u][i + j], f[lc[u]][i] + f[rc[u]][j] - (long long)i * j * 2 * a[u]);
      }
    }
    for (int i = (m); i >= (1); i--) {
      f[u][i] = max(f[u][i], f[u][i - 1] + (long long)(m - 1) * a[u] -
                                 (long long)(i - 1) * 2 * a[u]);
    }
  }
  int main() {
    scanf("%d%d", &n, &m);
    for (int i = (1); i <= (n); i++) scanf("%d", &a[i]);
    build();
    for (int i = (1); i <= (n); i++)
      for (int j = (1); j <= (m); j++) f[i][j] = -INF;
    dfs1(root);
    dfs2(root);
    for (int i = (1); i <= (n); i++) ans = max(ans, f[i][m]);
    printf("%lld\n", ans);
    return 0;
  }
  