#include <bits/stdc++.h>
using namespace std;
int a[4010], n, m, sta[4010], top, s[4010][2], sz[4010];
long long dp[4010][4010];
void DP(int x) {
  sz[x] = 1;
  for (int i = 0; i < 2; i++) {
    if (s[x][i]) {
      DP(s[x][i]);
      for (int t = min(m, sz[x]); t >= 0; t--)
        for (int j = min(sz[s[x][i]], m); j >= 1; j--)
          dp[x][t + j] = max(dp[x][t] + dp[s[x][i]][j] +
                                 1ll * (a[s[x][i]] - a[x]) * j * (m - j),
                             dp[x][t + j]);
      sz[x] += sz[s[x][i]];
    }
  }
  return;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n; i++) {
    while (top && a[sta[top]] > a[i]) s[i][0] = sta[top--];
    if (top) s[sta[top]][1] = i;
    sta[++top] = i;
  }
  int rt = 0;
  for (int i = 1; i <= n; i++)
    if (!rt || a[i] < a[rt]) rt = i;
  DP(rt);
  printf("%lld\n", dp[rt][m]);
}

### 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;
int a[4010], n, m, sta[4010], top, s[4010][2], sz[4010];
long long dp[4010][4010];
void DP(int x) {
  sz[x] = 1;
  for (int i = 0; i < 2; i++) {
    if (s[x][i]) {
      DP(s[x][i]);
      for (int t = min(m, sz[x]); t >= 0; t--)
        for (int j = min(sz[s[x][i]], m); j >= 1; j--)
          dp[x][t + j] = max(dp[x][t] + dp[s[x][i]][j] +
                                 1ll * (a[s[x][i]] - a[x]) * j * (m - j),
                             dp[x][t + j]);
      sz[x] += sz[s[x][i]];
    }
  }
  return;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n; i++) {
    while (top && a[sta[top]] > a[i]) s[i][0] = sta[top--];
    if (top) s[sta[top]][1] = i;
    sta[++top] = i;
  }
  int rt = 0;
  for (int i = 1; i <= n; i++)
    if (!rt || a[i] < a[rt]) rt = i;
  DP(rt);
  printf("%lld\n", dp[rt][m]);
}

```