#include <bits/stdc++.h>
using namespace std;
mt19937 rng(chrono::high_resolution_clock::now().time_since_epoch().count());
#pragma GCC target("avx2")
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
int n, m;
int a[4007];
int mn[4007][4007];
long long dp[4007][4007];
void f(int l, int r) {
  if (l >= r) {
    return;
  }
  int wh = mn[l][r];
  f(l, wh - 1);
  f(wh + 1, r);
  int tp = 0;
  for (int x = 0; x <= wh - l; ++x) {
    for (int y = 0; y <= r - wh; ++y) {
      long long aux = 0;
      aux += dp[mn[l][wh - 1]][x];
      if (l <= wh - 1) {
        aux += 1LL * (a[mn[l][wh - 1]] - a[mn[l][r]]) * x * (m - x);
      }
      aux += dp[mn[wh + 1][r]][y];
      if (wh + 1 <= r) {
        aux += 1LL * (a[mn[wh + 1][r]] - a[mn[l][r]]) * y * (m - y);
      }
      dp[wh][x + y] = max(dp[wh][x + y], aux);
    }
  }
  for (int i = r - l + 1; i >= 1; --i) {
    dp[wh][i] = max(dp[wh][i], dp[wh][i - 1]);
  }
}
void input() {
  cin >> n >> m;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
  }
  for (int i = 1; i <= n; ++i) {
    mn[i][i] = i;
    for (int j = i + 1; j <= n; ++j) {
      mn[i][j] = mn[i][j - 1];
      if (a[j] < a[mn[i][j]]) {
        mn[i][j] = j;
      }
    }
  }
}
void solve() {
  f(1, n);
  cout << dp[mn[1][n]][m] << "\n";
}
int main() {
  ios_base ::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  t = 1;
  while (t--) {
    input();
    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;
mt19937 rng(chrono::high_resolution_clock::now().time_since_epoch().count());
#pragma GCC target("avx2")
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
int n, m;
int a[4007];
int mn[4007][4007];
long long dp[4007][4007];
void f(int l, int r) {
  if (l >= r) {
    return;
  }
  int wh = mn[l][r];
  f(l, wh - 1);
  f(wh + 1, r);
  int tp = 0;
  for (int x = 0; x <= wh - l; ++x) {
    for (int y = 0; y <= r - wh; ++y) {
      long long aux = 0;
      aux += dp[mn[l][wh - 1]][x];
      if (l <= wh - 1) {
        aux += 1LL * (a[mn[l][wh - 1]] - a[mn[l][r]]) * x * (m - x);
      }
      aux += dp[mn[wh + 1][r]][y];
      if (wh + 1 <= r) {
        aux += 1LL * (a[mn[wh + 1][r]] - a[mn[l][r]]) * y * (m - y);
      }
      dp[wh][x + y] = max(dp[wh][x + y], aux);
    }
  }
  for (int i = r - l + 1; i >= 1; --i) {
    dp[wh][i] = max(dp[wh][i], dp[wh][i - 1]);
  }
}
void input() {
  cin >> n >> m;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
  }
  for (int i = 1; i <= n; ++i) {
    mn[i][i] = i;
    for (int j = i + 1; j <= n; ++j) {
      mn[i][j] = mn[i][j - 1];
      if (a[j] < a[mn[i][j]]) {
        mn[i][j] = j;
      }
    }
  }
}
void solve() {
  f(1, n);
  cout << dp[mn[1][n]][m] << "\n";
}
int main() {
  ios_base ::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  t = 1;
  while (t--) {
    input();
    solve();
  }
  return 0;
}

```