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 = 4005;
  using ll = long long;
  const ll MIN = LLONG_MIN >> 2;
  int n, m, a[N];
  inline ll calc(const vector<ll> &lans, const vector<ll> &rans, ll mn, int i,
                 int j) {
    return lans[i] + rans[j] - i * 1ll * j * 2 * mn;
  }
  vector<ll> solve(int l, int r) {
    auto ret = vector<ll>(m + 1, LLONG_MIN >> 2);
    ret[0] = 0;
    if (l > r) return ret;
    int mn = l;
    for (int i = l + 1; i <= r; i++)
      if (a[i] < a[mn]) mn = i;
    auto lans = solve(l, mn - 1), rans = solve(mn + 1, r);
    vector<ll> tt(m + 1, LLONG_MIN >> 2);
    int lim = min(m, r - l + 1);
    int i = 0, j = 0;
    tt[0] = 0;
    while (i + j < lim) {
      if (calc(lans, rans, a[mn], i, j) - calc(lans, rans, a[mn], i + 1, j) <
          calc(lans, rans, a[mn], i, j) - calc(lans, rans, a[mn], i, j + 1))
        ++i;
      else
        ++j;
      tt[i + j] = calc(lans, rans, a[mn], i, j);
    }
    for (int i = 0; i <= lim; i++) {
      ret[i] = max(ret[i], tt[i]);
      if (i) ret[i] = max(ret[i], tt[i - 1] + a[mn] * 1ll * (m - i * 2 + 1));
      if (i >= 2) assert(ret[i] - ret[i - 1] <= ret[i - 1] - ret[i - 2]);
    }
    return ret;
  }
  int main() {
    ios::sync_with_stdio(false);
    cin >> n >> m;
    for (int i = 1; i <= n; i++) cin >> a[i];
    cout << solve(1, n)[m] << endl;
    return 0;
  }
  