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 <typename T1, typename T2>
  void CMax(T1 &x, T2 y) {
    if (y > x) x = y;
  }
  template <typename T1, typename T2>
  void CMin(T1 &x, T2 y) {
    if (y < x) x = y;
  }
  template <typename T>
  void read(T &x) {
    x = 0;
    int op = 1;
    char c = getchar();
    while (!isdigit(c)) {
      if (c == '-') op = -1;
      c = getchar();
    }
    while (isdigit(c)) {
      x = (x << 1) + (x << 3) + (c ^ 48);
      c = getchar();
    }
    x *= op;
  }
  inline void swap(int &x, int &y) { x ^= y ^= x ^= y; }
  inline int max(int x, int y) { return x > y ? x : y; }
  inline int min(int x, int y) { return x < y ? x : y; }
  const int MN = 4005;
  int n, m, a[MN];
  int stk[MN], top;
  int ls[MN], rs[MN];
  int sz[MN];
  long long dp[MN][MN];
  void DP(int x) {
    sz[x] = 1;
    if (ls[x]) {
      DP(ls[x]);
      int w = a[ls[x]] - a[x];
      for (int i = min(m, sz[x]); ~i; --i)
        for (int j = min(m, sz[ls[x]]); ~j; --j)
          dp[x][i + j] =
              max(dp[x][i + j], dp[x][i] + dp[ls[x]][j] + 1ll * w * (m - j) * j);
      sz[x] += sz[ls[x]];
    }
    if (rs[x]) {
      DP(rs[x]);
      int w = a[rs[x]] - a[x];
      for (int i = min(m, sz[x]); ~i; --i)
        for (int j = min(m, sz[rs[x]]); ~j; --j)
          dp[x][i + j] =
              max(dp[x][i + j], dp[x][i] + dp[rs[x]][j] + 1ll * w * (m - j) * j);
      sz[x] += sz[rs[x]];
    }
  }
  int main() {
    read(n);
    read(m);
    for (int i = (1); i <= (n); ++i) read(a[i]);
    for (int i = 1; i <= n; ++i) {
      while (top && a[stk[top]] > a[i]) top--;
      if (!top)
        ls[i] = stk[1];
      else
        ls[i] = rs[stk[top]], rs[stk[top]] = i;
      stk[++top] = i;
    }
    DP(stk[1]);
    printf("%lld", dp[stk[1]][m]);
    return 0;
  }
  