#include <bits/stdc++.h>
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template <class T>
istream &operator>>(istream &in, vector<T> &v) {
  for (auto &x : v) {
    in >> x;
  }
  return in;
}
int main() {
  cin.tie(nullptr)->sync_with_stdio(false);
  int n, m;
  cin >> n >> m;
  vector<long long> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  function<vector<long long>(int, int)> dfs = [&](int l, int r) {
    if (l > r) {
      return vector<long long>{0};
    }
    int p = l;
    for (int i = l + 1; i <= r; i++) {
      if (a[i] < a[p]) {
        p = i;
      }
    }
    auto L = dfs(l, p - 1);
    auto R = dfs(p + 1, r);
    vector<long long> dp(L.size() + R.size() + 1);
    for (int i = 0; i < (int)L.size(); i++) {
      for (int j = 0; j < (int)R.size(); j++) {
        dp[i + j] = max(dp[i + j], L[i] + R[j] - (long long)a[p] * i * j * 2LL);
        dp[i + j + 1] =
            max(dp[i + j + 1], L[i] + R[j] + (long long)m * a[p] -
                                   a[p] * (2LL * (i + 1) * (j + 1) - 1));
      }
    }
    return dp;
  };
  cout << dfs(0, n - 1)[m] << "\n";
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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].

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template <class T>
istream &operator>>(istream &in, vector<T> &v) {
  for (auto &x : v) {
    in >> x;
  }
  return in;
}
int main() {
  cin.tie(nullptr)->sync_with_stdio(false);
  int n, m;
  cin >> n >> m;
  vector<long long> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  function<vector<long long>(int, int)> dfs = [&](int l, int r) {
    if (l > r) {
      return vector<long long>{0};
    }
    int p = l;
    for (int i = l + 1; i <= r; i++) {
      if (a[i] < a[p]) {
        p = i;
      }
    }
    auto L = dfs(l, p - 1);
    auto R = dfs(p + 1, r);
    vector<long long> dp(L.size() + R.size() + 1);
    for (int i = 0; i < (int)L.size(); i++) {
      for (int j = 0; j < (int)R.size(); j++) {
        dp[i + j] = max(dp[i + j], L[i] + R[j] - (long long)a[p] * i * j * 2LL);
        dp[i + j + 1] =
            max(dp[i + j + 1], L[i] + R[j] + (long long)m * a[p] -
                                   a[p] * (2LL * (i + 1) * (j + 1) - 1));
      }
    }
    return dp;
  };
  cout << dfs(0, n - 1)[m] << "\n";
}

```