#include <bits/stdc++.h>
#pragma GCC optimize("O3")
using namespace std;
const int maxn = 4003;
int n, m, val[maxn];
void read() {
  cin >> n >> m;
  for (int i = 1; i <= n; i++) cin >> val[i];
}
int lc[maxn], rc[maxn];
int f(int l, int r) {
  if (l > r) return -1;
  int min_val = INT_MAX, idx = -1;
  for (int i = l; i <= r; i++)
    if (val[i] < min_val) min_val = val[i], idx = i;
  lc[idx] = f(l, idx - 1), rc[idx] = f(idx + 1, r);
  return idx;
}
vector<long long> merge(vector<long long> a, vector<long long> b, int ver) {
  vector<long long> ans;
  ans.resize((int)a.size() + (int)b.size());
  for (int i = 0; i < (int)a.size(); i++)
    for (int j = 0; j < (int)b.size(); j++) {
      ans[i + j] = max(ans[i + j], a[i] + b[j] - 2ll * i * j * val[ver]);
      ans[i + j + 1] =
          max(ans[i + j + 1], a[i] + b[j] + (long long)(m - 1) * val[ver] -
                                  2ll * ((long long)i * j + i + j) * val[ver]);
    }
  return ans;
}
vector<long long> dfs(int ver) {
  if (ver == -1) return {0};
  return merge(dfs(lc[ver]), dfs(rc[ver]), ver);
}
void solve() {
  int root = f(1, n);
  auto ans = dfs(root);
  cout << ans[m] << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  read();
  solve();
}

### Prompt

Your challenge is to write 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>
#pragma GCC optimize("O3")
using namespace std;
const int maxn = 4003;
int n, m, val[maxn];
void read() {
  cin >> n >> m;
  for (int i = 1; i <= n; i++) cin >> val[i];
}
int lc[maxn], rc[maxn];
int f(int l, int r) {
  if (l > r) return -1;
  int min_val = INT_MAX, idx = -1;
  for (int i = l; i <= r; i++)
    if (val[i] < min_val) min_val = val[i], idx = i;
  lc[idx] = f(l, idx - 1), rc[idx] = f(idx + 1, r);
  return idx;
}
vector<long long> merge(vector<long long> a, vector<long long> b, int ver) {
  vector<long long> ans;
  ans.resize((int)a.size() + (int)b.size());
  for (int i = 0; i < (int)a.size(); i++)
    for (int j = 0; j < (int)b.size(); j++) {
      ans[i + j] = max(ans[i + j], a[i] + b[j] - 2ll * i * j * val[ver]);
      ans[i + j + 1] =
          max(ans[i + j + 1], a[i] + b[j] + (long long)(m - 1) * val[ver] -
                                  2ll * ((long long)i * j + i + j) * val[ver]);
    }
  return ans;
}
vector<long long> dfs(int ver) {
  if (ver == -1) return {0};
  return merge(dfs(lc[ver]), dfs(rc[ver]), ver);
}
void solve() {
  int root = f(1, n);
  auto ans = dfs(root);
  cout << ans[m] << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  read();
  solve();
}

```