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;
  std::mt19937 rnd(
      (int)std::chrono::steady_clock::now().time_since_epoch().count());
  long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
  const int MAXN = 4000;
  int n, ntake;
  int a[MAXN];
  bool taken[MAXN];
  long long cur[MAXN];
  long long solve() {
    for (int i = (0); i < (n); ++i) taken[i] = false;
    for (int i = (0); i < (n); ++i) cur[i] = (long long)(ntake - 1) * a[i];
    long long ret = 0;
    for (int i = (0); i < (ntake); ++i) {
      int best = -1;
      for (int j = (0); j < (n); ++j)
        if (!taken[j] && (best == -1 || cur[j] > cur[best])) best = j;
      ret += cur[best];
      taken[best] = true;
      {
        int lo = a[best];
        for (int j = best - 1; j >= 0; --j) {
          lo = min(lo, a[j]);
          cur[j] -= (long long)2 * lo;
        }
      }
      {
        int lo = a[best];
        for (int j = best + 1; j < n; ++j) {
          lo = min(lo, a[j]);
          cur[j] -= (long long)2 * lo;
        }
      }
    }
    return ret;
  }
  void run() {
    scanf("%d%d", &n, &ntake);
    for (int i = (0); i < (n); ++i) scanf("%d", &a[i]);
    printf("%lld\n", solve());
  }
  int main() {
    run();
    return 0;
  }
  