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;
  using ll = long long;
  constexpr int maxn = 4e3 + 7;
  int siz[maxn], ls[maxn], rs[maxn], lw[maxn], rw[maxn], arr[maxn];
  int sta[maxn], n, m;
  ll dp[maxn][maxn];
  void build() {
    int top = 0;
    for (int i = 1; i <= n; ++i) {
      while (top > 0 && arr[sta[top]] > arr[i]) ls[i] = sta[top--];
      if (sta[top]) {
        rs[sta[top]] = i;
        rw[sta[top]] = arr[i] - arr[sta[top]];
      }
      if (ls[i]) {
        lw[i] = arr[ls[i]] - arr[i];
      }
      sta[++top] = i;
    }
  }
  void dfs(int u) {
    siz[u] = 1;
    if (ls[u]) {
      dfs(ls[u]);
      for (int i = min(m, siz[u]); i >= 0; --i) {
        for (int j = min(m, siz[ls[u]]); j >= 0; --j) {
          if (i + j > m) continue;
          dp[u][i + j] = max(dp[u][i + j],
                             dp[u][i] + dp[ls[u]][j] + 1ll * (m - j) * j * lw[u]);
        }
      }
      siz[u] += siz[ls[u]];
    }
    if (rs[u]) {
      dfs(rs[u]);
      for (int i = min(m, siz[u]); i >= 0; --i) {
        for (int j = min(m, siz[rs[u]]); j >= 0; --j) {
          if (i + j > m) continue;
          dp[u][i + j] = max(dp[u][i + j],
                             dp[u][i] + dp[rs[u]][j] + 1ll * (m - j) * j * rw[u]);
        }
      }
      siz[u] += siz[rs[u]];
    }
  }
  void solve() {
    scanf("%d %d", &n, &m);
    for (int i = 1; i <= n; ++i) scanf("%d", arr + i);
    build();
    dfs(sta[1]);
    printf("%lld\n", dp[sta[1]][m]);
  }
  int main() {
    solve();
    return 0;
  }
  