#include <bits/stdc++.h>
using namespace std;
const int nax = 4005;
int n, m;
long long int a[nax];
vector<long long int> reku(int L, int R) {
  if (L > R) return {0};
  if (L == R) {
    return {0, a[L] * (m - 1)};
  }
  int pos = L;
  long long int v = a[L];
  for (int i = L; i <= R; i++) {
    if (a[i] < v) {
      v = a[i];
      pos = i;
    }
  }
  auto LF = reku(L, pos - 1);
  auto RG = reku(pos + 1, R);
  vector<long long int> ans(R - L + 2, -2e18);
  for (int i = 0; i < LF.size(); i++) {
    for (int j = 0; j < RG.size(); j++) {
      ans[i + j] = max(ans[i + j], LF[i] + RG[j] - v * i * j * 2);
      ans[i + j + 1] =
          max(ans[i + j + 1],
              LF[i] + RG[j] + v * m - v * i * j * 2 - v - v * (i + j) * 2);
    }
  }
  return ans;
}
void solve() {
  cin >> n >> m;
  for (int i = 1; i <= n; i++) cin >> a[i];
  auto res = reku(1, n);
  cout << res[m] << "\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int tt = 1;
  while (tt--) solve();
  return 0;
}

### Prompt

Please create a solution in cpp 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 nax = 4005;
int n, m;
long long int a[nax];
vector<long long int> reku(int L, int R) {
  if (L > R) return {0};
  if (L == R) {
    return {0, a[L] * (m - 1)};
  }
  int pos = L;
  long long int v = a[L];
  for (int i = L; i <= R; i++) {
    if (a[i] < v) {
      v = a[i];
      pos = i;
    }
  }
  auto LF = reku(L, pos - 1);
  auto RG = reku(pos + 1, R);
  vector<long long int> ans(R - L + 2, -2e18);
  for (int i = 0; i < LF.size(); i++) {
    for (int j = 0; j < RG.size(); j++) {
      ans[i + j] = max(ans[i + j], LF[i] + RG[j] - v * i * j * 2);
      ans[i + j + 1] =
          max(ans[i + j + 1],
              LF[i] + RG[j] + v * m - v * i * j * 2 - v - v * (i + j) * 2);
    }
  }
  return ans;
}
void solve() {
  cin >> n >> m;
  for (int i = 1; i <= n; i++) cin >> a[i];
  auto res = reku(1, n);
  cout << res[m] << "\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int tt = 1;
  while (tt--) solve();
  return 0;
}

```