#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const ll INF = 1e18;
const int N = 4000;
vector<ll> vals[N + 1];
int pre[N];
int nxt[N];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  vector<ll> as(n);
  for (ll& a : as) cin >> a;
  vector<pair<ll, int>> ord(n);
  for (int i = 0; i < n; ++i) {
    ord[i] = {as[i], i};
    pre[i] = i;
  }
  sort(ord.rbegin(), ord.rend());
  for (auto pr : ord) {
    int i = pr.second;
    vector<ll> le, ri;
    int lf = i, rf = i;
    if (i > 0 && as[i - 1] > as[i]) {
      swap(le, vals[pre[i - 1]]);
      lf = pre[i - 1];
    } else {
      le.push_back(0);
    }
    if (i + 1 < n && as[i + 1] > as[i]) {
      swap(ri, vals[i + 1]);
      rf = nxt[i + 1];
    } else {
      ri.push_back(0);
    }
    pre[rf] = lf;
    nxt[lf] = rf;
    vals[lf].resize((int)le.size() + (int)ri.size(), -INF);
    for (int x = 0; x < le.size(); ++x) {
      for (int y = 0; y < ri.size(); ++y) {
        ll off0 = le[x] + ri[y] - 2 * x * y * as[i];
        ll off1 = le[x] + ri[y] + as[i] * (m - 2 * x * y - 2 * x - 2 * y - 1);
        vals[lf][x + y] = max(vals[lf][x + y], off0);
        vals[lf][x + y + 1] = max(vals[lf][x + y + 1], off1);
      }
    }
  }
  cout << vals[0][m] << '\n';
}

### Prompt

Your task is to create a cpp solution to the following problem:

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;
using ll = long long;
const ll INF = 1e18;
const int N = 4000;
vector<ll> vals[N + 1];
int pre[N];
int nxt[N];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  vector<ll> as(n);
  for (ll& a : as) cin >> a;
  vector<pair<ll, int>> ord(n);
  for (int i = 0; i < n; ++i) {
    ord[i] = {as[i], i};
    pre[i] = i;
  }
  sort(ord.rbegin(), ord.rend());
  for (auto pr : ord) {
    int i = pr.second;
    vector<ll> le, ri;
    int lf = i, rf = i;
    if (i > 0 && as[i - 1] > as[i]) {
      swap(le, vals[pre[i - 1]]);
      lf = pre[i - 1];
    } else {
      le.push_back(0);
    }
    if (i + 1 < n && as[i + 1] > as[i]) {
      swap(ri, vals[i + 1]);
      rf = nxt[i + 1];
    } else {
      ri.push_back(0);
    }
    pre[rf] = lf;
    nxt[lf] = rf;
    vals[lf].resize((int)le.size() + (int)ri.size(), -INF);
    for (int x = 0; x < le.size(); ++x) {
      for (int y = 0; y < ri.size(); ++y) {
        ll off0 = le[x] + ri[y] - 2 * x * y * as[i];
        ll off1 = le[x] + ri[y] + as[i] * (m - 2 * x * y - 2 * x - 2 * y - 1);
        vals[lf][x + y] = max(vals[lf][x + y], off0);
        vals[lf][x + y + 1] = max(vals[lf][x + y + 1], off1);
      }
    }
  }
  cout << vals[0][m] << '\n';
}

```