#include <bits/stdc++.h>
using namespace std;
const int MOD = 998244353;
template <class T>
using vi2 = vector<vector<T>>;
using ll = long long;
using pii = pair<int, int>;
int main() {
  auto solve = [&]() {
    int n, m;
    cin >> n >> m;
    vector<int> a(n + 1), per(n + 1), fa(n + 1), sz(n + 1);
    for (int i = 1, _i = int(n); i <= _i; i++) cin >> a[i];
    function<int(int)> root = [&](int x) {
      return fa[x] ? fa[x] = root(fa[x]) : x;
    };
    auto un = [&](int a, int b) {
      a = root(a), b = root(b);
      fa[a] = b;
    };
    for (int i = 1, _i = int(n); i <= _i; i++) per[i] = i;
    sort(per.begin() + 1, per.end(), [&](int x, int y) { return a[x] > a[y]; });
    vi2<int> g(n + 1);
    for (int i = 1, _i = int(n); i <= _i; i++) {
      int x = per[i];
      if (x != 1 && a[x - 1] > a[x]) {
        int r = root(x - 1);
        g[x].emplace_back(r);
        un(r, x);
      }
      if (x != n && a[x + 1] > a[x]) {
        int r = root(x + 1);
        g[x].emplace_back(r);
        un(r, x);
      }
    }
    int r = per[n];
    vi2<ll> dp(n + 1);
    function<void(int, int)> dfs = [&](int x, int fa) {
      vector<ll> dp1(2, -LLONG_MAX / 3);
      dp1[0] = 0;
      dp1[1] = 1LL * (m - 1) * a[x];
      sz[x] = 1;
      for (int to : g[x]) {
        dfs(to, x);
        vector<ll> tmp1(sz[x] + sz[to] + 1, -LLONG_MAX / 3);
        for (int l = 0, _l = int(sz[x]); l <= _l; l++)
          for (int r = 0, _r = int(sz[to]); r <= _r; r++) {
            tmp1[l + r] =
                max(dp1[l] + dp[to][r] - 2LL * l * r * a[x], tmp1[l + r]);
          }
        sz[x] += sz[to];
        dp1 = tmp1;
      }
      dp[x] = dp1;
    };
    dfs(r, r);
    cout << dp[r][m] << endl;
  };
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

Generate 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;
const int MOD = 998244353;
template <class T>
using vi2 = vector<vector<T>>;
using ll = long long;
using pii = pair<int, int>;
int main() {
  auto solve = [&]() {
    int n, m;
    cin >> n >> m;
    vector<int> a(n + 1), per(n + 1), fa(n + 1), sz(n + 1);
    for (int i = 1, _i = int(n); i <= _i; i++) cin >> a[i];
    function<int(int)> root = [&](int x) {
      return fa[x] ? fa[x] = root(fa[x]) : x;
    };
    auto un = [&](int a, int b) {
      a = root(a), b = root(b);
      fa[a] = b;
    };
    for (int i = 1, _i = int(n); i <= _i; i++) per[i] = i;
    sort(per.begin() + 1, per.end(), [&](int x, int y) { return a[x] > a[y]; });
    vi2<int> g(n + 1);
    for (int i = 1, _i = int(n); i <= _i; i++) {
      int x = per[i];
      if (x != 1 && a[x - 1] > a[x]) {
        int r = root(x - 1);
        g[x].emplace_back(r);
        un(r, x);
      }
      if (x != n && a[x + 1] > a[x]) {
        int r = root(x + 1);
        g[x].emplace_back(r);
        un(r, x);
      }
    }
    int r = per[n];
    vi2<ll> dp(n + 1);
    function<void(int, int)> dfs = [&](int x, int fa) {
      vector<ll> dp1(2, -LLONG_MAX / 3);
      dp1[0] = 0;
      dp1[1] = 1LL * (m - 1) * a[x];
      sz[x] = 1;
      for (int to : g[x]) {
        dfs(to, x);
        vector<ll> tmp1(sz[x] + sz[to] + 1, -LLONG_MAX / 3);
        for (int l = 0, _l = int(sz[x]); l <= _l; l++)
          for (int r = 0, _r = int(sz[to]); r <= _r; r++) {
            tmp1[l + r] =
                max(dp1[l] + dp[to][r] - 2LL * l * r * a[x], tmp1[l + r]);
          }
        sz[x] += sz[to];
        dp1 = tmp1;
      }
      dp[x] = dp1;
    };
    dfs(r, r);
    cout << dp[r][m] << endl;
  };
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

```