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 = 4010;
  const int inf = 2147483647;
  long long read() {
    long long x = 0, f = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9') {
      if (ch == '-') f = -1;
      ch = getchar();
    }
    while (ch >= '0' && ch <= '9') x = x * 10ll + ch - '0', ch = getchar();
    return x * f;
  }
  int n, m, sz[N];
  long long a[N], f[N][N];
  int solve(int l, int r) {
    if (l > r) return 0;
    int p = l;
    for (int i = l + 1; i <= r; i++)
      if (a[i] < a[p]) p = i;
    sz[p] = r - l + 1;
    int L = solve(l, p - 1), R = solve(p + 1, r);
    for (int i = 0; i <= sz[L]; i++)
      for (int j = 0; j <= sz[R]; j++)
        f[p][i + j] = max(f[p][i + j], f[L][i] + f[R][j] - 2ll * i * j * a[p]),
                 f[p][i + j + 1] = max(f[p][i + j + 1],
                                       f[L][i] + f[R][j] + (long long)m * a[p] -
                                           2ll * (i + 1) * (j + 1) * a[p] + a[p]);
    return p;
  }
  int main() {
    n = read(), m = read();
    for (int i = 1; i <= n; i++) a[i] = read();
    printf("%lld\n", f[solve(1, n)][m]);
    return 0;
  }
  