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;
  template <class T>
  inline void cmin(T &a, T b) {
    ((a > b) && (a = b));
  }
  template <class T>
  inline void cmax(T &a, T b) {
    ((a < b) && (a = b));
  }
  char IO;
  template <class T = int>
  T rd() {
    T s = 0;
    int f = 0;
    while (!isdigit(IO = getchar())) f |= IO == '-';
    do s = (s << 1) + (s << 3) + (IO ^ '0');
    while (isdigit(IO = getchar()));
    return f ? -s : s;
  }
  bool Mbe;
  const int N = 4010, INF = 1e9 + 10;
  int n, m;
  long long A[N], dp[N][N];
  int Go(int l, int r) {
    if (l > r) return 0;
    int u = l;
    for (int i = l, iend = r; i <= iend; ++i)
      if (A[i] < A[u]) u = i;
    for (int i = 1, iend = r - l + 1; i <= iend; ++i) dp[u][i] = -1e18;
    int ls = Go(l, u - 1);
    int rs = Go(u + 1, r);
    for (int i = 0, iend = u - l; i <= iend; ++i)
      for (int j = 0, jend = r - u; j <= jend; ++j) {
        cmax(dp[u][i + j], dp[ls][i] + dp[rs][j] - 2ll * A[u] * i * j);
        cmax(dp[u][i + j + 1],
             dp[ls][i] + dp[rs][j] - A[u] * (2 * (i * j + i + j) + 1 - m));
      }
    return u;
  }
  bool Med;
  int main() {
    n = rd(), m = rd();
    for (int i = 1, iend = n; i <= iend; ++i) A[i] = rd();
    int u = Go(1, n);
    printf("%lld\n", dp[u][m]);
  }
  