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, ch = getchar();
    while (ch < '0' || ch > '9') {
      if (ch == '-') f = -1;
      ch = getchar();
    }
    while (ch >= '0' && ch <= '9') {
      x = x * 10 + ch - '0';
      ch = getchar();
    }
    return x * f;
  }
  inline void write(long long x) {
    if (x < 0) putchar('-'), x = -x;
    if (x >= 10) write(x / 10);
    putchar(x % 10 + '0');
  }
  inline void writeln(long long x) {
    write(x);
    puts("");
  }
  const int N = 4005;
  const long long INF = 1e15;
  int n, m, mn[N][N], tot;
  long long a[N], f[N][N];
  short int mp[N][N];
  bool g[N][N];
  inline void init() {
    n = read();
    m = read();
    for (int i = 1; i <= n; i++) a[i] = read();
  }
  inline long long dp(int l, int r, int k) {
    if (k < 0) return -INF;
    if (l > r) {
      if (!k) return 0;
      return -INF;
    }
    if (l == r) {
      if (k == 0) return 0;
      if (k == 1) return 1ll * a[l] * (m - 1);
      return -INF;
    }
    if (!mp[l][r]) mp[l][r] = ++tot;
    int x = mp[l][r];
    if (g[x][k]) return f[x][k];
    g[x][k] = 1;
    int p = mn[l][r];
    if (p == l)
      return f[x][k] =
                 max(dp(l + 1, r, k), dp(l + 1, r, k - 1) + a[l] * (m - 1ll) -
                                          2ll * a[l] * (k - 1));
    if (p == r)
      return f[x][k] =
                 max(dp(l, r - 1, k), dp(l, r - 1, k - 1) + a[r] * (m - 1ll) -
                                          2ll * a[r] * (k - 1));
    if (p - l < r - p) {
      for (int i = 0; i <= min(k, p - l + 1); i++) {
        f[x][k] = max(f[x][k], dp(l, p, i) + dp(p + 1, r, k - i) -
                                   2ll * (k - i) * i * a[p]);
      }
    } else {
      for (int i = 0; i <= min(k, r - p + 1); i++) {
        f[x][k] = max(f[x][k], dp(l, p, k - i) + dp(p + 1, r, i) -
                                   2ll * (k - i) * i * a[p]);
      }
    }
    return f[x][k];
  }
  inline void solve() {
    for (int i = 1; i <= n; i++) {
      mn[i][i] = i;
      for (int j = i + 1; j <= n; j++) {
        if (a[j] < a[mn[i][j - 1]])
          mn[i][j] = j;
        else
          mn[i][j] = mn[i][j - 1];
      }
    }
    writeln(dp(1, n, m));
  }
  int main() {
    init();
    solve();
    return 0;
  }
  