#include <bits/stdc++.h>
using namespace std;
mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
mt19937 rnf(2106);
const int N = 4003;
const long long INF = 1000000009000000009;
int n, m;
long long a[N];
int mini[N][N];
long long dp[N][N];
void dfs(int l, int r) {
  if (l > r) return;
  int x = mini[l][r];
  dfs(l, x - 1);
  dfs(x + 1, r);
  int hl = mini[l][x - 1];
  int hr = mini[x + 1][r];
  int ql = (x - l);
  int qr = (r - x);
  for (int i = 0; i <= ql; ++i) {
    for (int j = 0; j <= qr; ++j) {
      long long s = dp[hl][i] + dp[hr][j];
      s -= i * j * a[x] * 2;
      dp[x][i + j] = max(dp[x][i + j], s);
      s += m * a[x];
      s += i * j * a[x] * 2;
      s -= (i + 1) * (j + 1) * a[x] * 2;
      s += a[x];
      dp[x][i + j + 1] = max(dp[x][i + j + 1], s);
    }
  }
}
void solv() {
  cin >> n >> m;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  for (int i = 1; i <= n; ++i) {
    mini[i][i] = i;
    for (int j = i + 1; j <= n; ++j) {
      if (a[j] < a[mini[i][j - 1]])
        mini[i][j] = j;
      else
        mini[i][j] = mini[i][j - 1];
    }
  }
  for (int x = 0; x <= n; ++x) {
    for (int i = 0; i <= n; ++i) {
      dp[x][i] = -INF;
    }
  }
  dp[0][0] = 0;
  dfs(1, n);
  cout << dp[mini[1][n]][m] << "\n";
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0);
  int tt = 1;
  while (tt--) solv();
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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 rnd(chrono::steady_clock::now().time_since_epoch().count());
mt19937 rnf(2106);
const int N = 4003;
const long long INF = 1000000009000000009;
int n, m;
long long a[N];
int mini[N][N];
long long dp[N][N];
void dfs(int l, int r) {
  if (l > r) return;
  int x = mini[l][r];
  dfs(l, x - 1);
  dfs(x + 1, r);
  int hl = mini[l][x - 1];
  int hr = mini[x + 1][r];
  int ql = (x - l);
  int qr = (r - x);
  for (int i = 0; i <= ql; ++i) {
    for (int j = 0; j <= qr; ++j) {
      long long s = dp[hl][i] + dp[hr][j];
      s -= i * j * a[x] * 2;
      dp[x][i + j] = max(dp[x][i + j], s);
      s += m * a[x];
      s += i * j * a[x] * 2;
      s -= (i + 1) * (j + 1) * a[x] * 2;
      s += a[x];
      dp[x][i + j + 1] = max(dp[x][i + j + 1], s);
    }
  }
}
void solv() {
  cin >> n >> m;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  for (int i = 1; i <= n; ++i) {
    mini[i][i] = i;
    for (int j = i + 1; j <= n; ++j) {
      if (a[j] < a[mini[i][j - 1]])
        mini[i][j] = j;
      else
        mini[i][j] = mini[i][j - 1];
    }
  }
  for (int x = 0; x <= n; ++x) {
    for (int i = 0; i <= n; ++i) {
      dp[x][i] = -INF;
    }
  }
  dp[0][0] = 0;
  dfs(1, n);
  cout << dp[mini[1][n]][m] << "\n";
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0);
  int tt = 1;
  while (tt--) solv();
  return 0;
}

```