#include <bits/stdc++.h>
using namespace std;
int n, m, rt, sz[4005];
int st[4005], top, ch[4005][2];
long long dp[4005][4005], a[4005];
void dfs(int now) {
  sz[now] = 1;
  for (int i = 0; i < 2; i++)
    if (ch[now][i]) {
      int to = ch[now][i];
      dfs(to);
      for (int j = sz[now]; j >= 0; j--)
        for (int k = sz[to]; k >= 0; k--)
          dp[now][j + k] =
              max(dp[now][j + k],
                  dp[now][j] + dp[to][k] + k * (m - k) * (a[to] - a[now]));
      sz[now] += sz[to];
    }
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
    while (top && a[st[top]] > a[i]) ch[i][0] = st[top--];
    if (top) ch[st[top]][1] = i;
    st[++top] = i;
  }
  for (int i = 1; i <= n; i++)
    if (!rt || a[i] < a[rt]) rt = i;
  dfs(rt);
  printf("%lld", dp[rt][m]);
}

### Prompt

Please provide a cpp coded solution 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;
int n, m, rt, sz[4005];
int st[4005], top, ch[4005][2];
long long dp[4005][4005], a[4005];
void dfs(int now) {
  sz[now] = 1;
  for (int i = 0; i < 2; i++)
    if (ch[now][i]) {
      int to = ch[now][i];
      dfs(to);
      for (int j = sz[now]; j >= 0; j--)
        for (int k = sz[to]; k >= 0; k--)
          dp[now][j + k] =
              max(dp[now][j + k],
                  dp[now][j] + dp[to][k] + k * (m - k) * (a[to] - a[now]));
      sz[now] += sz[to];
    }
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
    while (top && a[st[top]] > a[i]) ch[i][0] = st[top--];
    if (top) ch[st[top]][1] = i;
    st[++top] = i;
  }
  for (int i = 1; i <= n; i++)
    if (!rt || a[i] < a[rt]) rt = i;
  dfs(rt);
  printf("%lld", dp[rt][m]);
}

```