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;
  long long read() {
    long long x = 0, b = 1;
    char c;
    do c = getchar();
    while ((c > '9' || c < '0') && c != '-');
    if (c == '-') b = 0;
    do x = x * 10 + c - '0', c = getchar();
    while (c >= '0' && c <= '9');
    return b ? x : -x;
  }
  const long long maxn = 4010;
  long long i, j, k, n, m, T, a[maxn], son[maxn][2], root, cnt, dp[maxn][maxn],
      size[maxn];
  void dfs(long long &root, long long L, long long R) {
    if (L > R) return;
    if (!root) root = ++cnt;
    long long Min = L;
    for (long long i = L; i <= R; i++)
      if (a[i] < a[Min]) Min = i;
    dfs(son[root][0], L, Min - 1);
    dfs(son[root][1], Min + 1, R);
    size[root] = size[son[root][0]] + size[son[root][1]] + 1;
    for (long long i = 0; i <= size[son[root][0]]; i++)
      for (long long j = 0; j <= size[son[root][1]]; j++) {
        dp[root][i + j] =
            max(dp[root][i + j],
                dp[son[root][0]][i] + dp[son[root][1]][j] - 2 * i * j * a[Min]);
        dp[root][i + j + 1] =
            max(dp[root][i + j + 1],
                dp[son[root][0]][i] + dp[son[root][1]][j] + m * a[Min] -
                    (2 * i * j + 2 * i + 2 * j + 1) * a[Min]);
      }
  }
  signed main() {
    cin >> n >> m;
    for (i = 1; i <= n; i++) a[i] = read();
    dfs(root, 1, n);
    long long Ans = 0;
    cout << dp[root][m] << endl;
  }
  