#include <bits/stdc++.h>
using namespace std;
const int maxn = 4000 + 10;
const long long INF = 4e18;
int m, a[maxn];
int newInt = 1;
long long dp[maxn][maxn];
int sz[maxn];
int solve(int l, int r) {
  if (l == r) {
    dp[0][0] = 0;
    return 0;
  }
  int v = newInt++;
  dp[v][0] = 0;
  if (l + 1 == r) {
    dp[v][1] = 1LL * (m - 1) * a[l];
    sz[v] = 1;
    return v;
  }
  int idx = l;
  for (int i = l; i < r; i++)
    if (a[i] < a[idx]) idx = i;
  int lc = solve(l, idx);
  int rc = solve(idx + 1, r);
  int sz1 = sz[lc], sz2 = sz[rc];
  for (int i = 1; i <= sz1 + 1; i++)
    dp[v][i] = max(dp[lc][i], dp[lc][i - 1] + 1LL * (m - (2 * i - 1)) * a[idx]);
  for (int i = sz1 + 1; i >= 0; i--)
    for (int j = sz2; j >= 0; j--)
      dp[v][i + j] =
          max(dp[v][i + j], dp[v][i] + dp[rc][j] - 2LL * i * j * a[idx]);
  sz[v] = sz1 + sz2 + 1;
  return v;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n >> m;
  for (int i = 1; i <= n; i++) cin >> a[i];
  memset(dp, -63, sizeof dp);
  solve(1, n + 1);
  cout << dp[1][m] << '\n';
}

### Prompt

Please formulate 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 maxn = 4000 + 10;
const long long INF = 4e18;
int m, a[maxn];
int newInt = 1;
long long dp[maxn][maxn];
int sz[maxn];
int solve(int l, int r) {
  if (l == r) {
    dp[0][0] = 0;
    return 0;
  }
  int v = newInt++;
  dp[v][0] = 0;
  if (l + 1 == r) {
    dp[v][1] = 1LL * (m - 1) * a[l];
    sz[v] = 1;
    return v;
  }
  int idx = l;
  for (int i = l; i < r; i++)
    if (a[i] < a[idx]) idx = i;
  int lc = solve(l, idx);
  int rc = solve(idx + 1, r);
  int sz1 = sz[lc], sz2 = sz[rc];
  for (int i = 1; i <= sz1 + 1; i++)
    dp[v][i] = max(dp[lc][i], dp[lc][i - 1] + 1LL * (m - (2 * i - 1)) * a[idx]);
  for (int i = sz1 + 1; i >= 0; i--)
    for (int j = sz2; j >= 0; j--)
      dp[v][i + j] =
          max(dp[v][i + j], dp[v][i] + dp[rc][j] - 2LL * i * j * a[idx]);
  sz[v] = sz1 + sz2 + 1;
  return v;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n >> m;
  for (int i = 1; i <= n; i++) cin >> a[i];
  memset(dp, -63, sizeof dp);
  solve(1, n + 1);
  cout << dp[1][m] << '\n';
}

```