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;
  inline long long read() {
    long long x = 0, f = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9') f = ch == '-' ? -1 : 1, ch = getchar();
    while (ch >= '0' && ch <= '9') x = x * 10 + ch - 48, ch = getchar();
    return x * f;
  }
  const long long N(4e3 + 5);
  long long rt;
  long long top;
  long long n, m;
  long long a[N];
  long long stk[N];
  long long siz[N];
  long long w[N][2];
  long long ch[N][2];
  long long dp[N][N];
  inline void dfs(long long u) {
    siz[u] = 1;
    for (long long c = (0); c <= (1); c++)
      if (ch[u][c]) {
        long long v = ch[u][c];
        dfs(v);
        for (long long i = (siz[u]); i >= (0); i--)
          for (long long j = (siz[v]); j >= (0); j--)
            ((dp[u][i + j]) = max((dp[u][i + j]),
                                  (dp[u][i] + dp[v][j] + j * (m - j) * w[u][c])));
        siz[u] += siz[v];
      }
  }
  signed main() {
    n = read(), m = read();
    for (long long i = (1); i <= (n); i++) a[i] = read();
    rt = 1;
    for (long long i = (1); i <= (n); i++) {
      long long k = top;
      if (a[i] < a[rt]) rt = i;
      while (k && a[stk[k]] > a[i]) k--;
      if (k) ch[stk[k]][1] = i, w[stk[k]][1] = a[i] - a[stk[k]];
      if (k < top) ch[i][0] = stk[k + 1], w[i][0] = a[stk[k + 1]] - a[i];
      stk[top = (++k)] = i;
    }
    dfs(rt), cout << dp[rt][m];
    return 0;
  }
  