#include <bits/stdc++.h>
using namespace std;
void debug() { cout << endl; }
template <class T, class... U>
void debug(T a, U... b) {
  cout << a << " ";
  debug(b...);
}
template <class T>
void pary(T l, T r) {
  while (l != r) {
    cout << *l << " ";
    l++;
  }
  cout << endl;
}
const long long inf = (1LL << 60);
vector<pair<int, int> > adj[4005];
int a[4005], par[4005];
int m;
vector<long long> dfs(int n, int pa) {
  vector<long long> ret(2, 0);
  for (auto v : adj[n]) {
    if (v.first != pa) {
      vector<long long> tmp = dfs(v.first, n);
      int siz = ret.size() + tmp.size() - 1;
      vector<long long> cur(siz, -inf);
      for (int i = 0; i < siz; i++) {
        long long val = -inf;
        for (int j = max(i - (int)ret.size() + 1, 0);
             j <= min(i, (int)tmp.size() - 1); j++) {
          val =
              max(val, tmp[j] + (long long)v.second * j * (m - j) + ret[i - j]);
        }
        cur[i] = val;
      }
      ret = cur;
    }
  }
  return ret;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n >> m;
  for (int i = 1; i <= n; i++) cin >> a[i];
  stack<int> stk;
  for (int i = 1; i <= n; i++) {
    int prv = 0;
    while (stk.size() && a[i] <= a[stk.top()]) {
      prv = stk.top();
      stk.pop();
    }
    if (prv) par[prv] = i;
    if (stk.size()) par[i] = stk.top();
    stk.push(i);
  }
  for (int i = 1; i <= n; i++) {
    if (par[i]) {
      adj[i].push_back({par[i], a[i] - a[par[i]]});
      adj[par[i]].push_back({i, a[i] - a[par[i]]});
    }
  }
  vector<long long> ans = dfs(1, 0);
  cout << ans[m] << "\n";
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
void debug() { cout << endl; }
template <class T, class... U>
void debug(T a, U... b) {
  cout << a << " ";
  debug(b...);
}
template <class T>
void pary(T l, T r) {
  while (l != r) {
    cout << *l << " ";
    l++;
  }
  cout << endl;
}
const long long inf = (1LL << 60);
vector<pair<int, int> > adj[4005];
int a[4005], par[4005];
int m;
vector<long long> dfs(int n, int pa) {
  vector<long long> ret(2, 0);
  for (auto v : adj[n]) {
    if (v.first != pa) {
      vector<long long> tmp = dfs(v.first, n);
      int siz = ret.size() + tmp.size() - 1;
      vector<long long> cur(siz, -inf);
      for (int i = 0; i < siz; i++) {
        long long val = -inf;
        for (int j = max(i - (int)ret.size() + 1, 0);
             j <= min(i, (int)tmp.size() - 1); j++) {
          val =
              max(val, tmp[j] + (long long)v.second * j * (m - j) + ret[i - j]);
        }
        cur[i] = val;
      }
      ret = cur;
    }
  }
  return ret;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n >> m;
  for (int i = 1; i <= n; i++) cin >> a[i];
  stack<int> stk;
  for (int i = 1; i <= n; i++) {
    int prv = 0;
    while (stk.size() && a[i] <= a[stk.top()]) {
      prv = stk.top();
      stk.pop();
    }
    if (prv) par[prv] = i;
    if (stk.size()) par[i] = stk.top();
    stk.push(i);
  }
  for (int i = 1; i <= n; i++) {
    if (par[i]) {
      adj[i].push_back({par[i], a[i] - a[par[i]]});
      adj[par[i]].push_back({i, a[i] - a[par[i]]});
    }
  }
  vector<long long> ans = dfs(1, 0);
  cout << ans[m] << "\n";
}

```