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 int N = 4005;
  const long long INF = 0x3f3f3f3f3f3f3f3f;
  template <typename T>
  inline void read(T &WOW) {
    T x = 0, flag = 1;
    char ch = getchar();
    while (!isdigit(ch)) {
      if (ch == '-') flag = -1;
      ch = getchar();
    }
    while (isdigit(ch)) {
      x = x * 10 + ch - '0';
      ch = getchar();
    }
    WOW = flag * x;
  }
  template <typename T>
  inline void cmax(T &x, T y) {
    if (x < y) x = y;
  }
  int n, m, siz[N];
  long long a[N], dp[N][N];
  int DP(int l, int r) {
    if (l > r) return 0;
    if (l == r) {
      siz[l] = 1;
      dp[l][0] = 0;
      dp[l][1] = a[l] * (m - 1);
      return l;
    }
    int u = l;
    for (int i = l + 1; i <= r; ++i) {
      if (a[i] < a[u]) u = i;
    }
    int v1 = DP(l, u - 1), v2 = DP(u + 1, r);
    siz[u] = min(m, r - l + 1);
    for (int i = 0; i <= siz[u]; ++i) {
      dp[u][i] = -INF;
    }
    for (int i = 0; i <= siz[v1]; ++i) {
      for (int j = 0; j <= siz[v2] && i + j <= siz[u]; ++j) {
        cmax(dp[u][i + j], dp[v1][i] + dp[v2][j] - a[u] * 2 * i * j);
        if (i + j + 1 <= siz[u]) {
          cmax(dp[u][i + j + 1],
               dp[v1][i] + dp[v2][j] - a[u] * (2 * (i * j + i + j) - (m - 1)));
        }
      }
    }
    return u;
  }
  void solve() {
    read(n);
    read(m);
    for (int i = 1; i <= n; ++i) {
      read(a[i]);
    }
    int rt = DP(1, n);
    printf("%lld\n", dp[rt][m]);
  }
  int main() {
    solve();
    return 0;
  }
  