#include <bits/stdc++.h>
using namespace std;
int n, m;
int a[4086];
int son[4086][2];
int sk[4086], top;
long long f[4086][4086], g[4086];
int siz[4086];
void dfs(int i) {
  f[i][0] = f[i][1] = 0, siz[i] = 1;
  for (int j = 0; j <= 1; j++) {
    int to = son[i][j];
    if (!to) continue;
    dfs(to);
    for (int k = 0; k <= min(siz[i], m); k++) {
      for (int l = 0; l <= min(siz[to], m - k); l++) {
        g[k + l] = max(g[k + l],
                       f[i][k] + f[to][l] + 1ll * (a[to] - a[i]) * l * (m - l));
      }
    }
    for (int k = 0; k <= min(siz[i] + siz[to], m); k++)
      f[i][k] = g[k], g[k] = -1e18;
    siz[i] += siz[to];
  }
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    int last = 0;
    while (top && a[sk[top]] > a[i]) last = sk[top--];
    son[i][0] = last, son[sk[top]][1] = i;
    sk[++top] = i;
  }
  memset(f, -0x3f, sizeof(f)), memset(g, -0x3f, sizeof(g));
  dfs(sk[1]);
  printf("%lld", f[sk[1]][m]);
}

### Prompt

Your task is to create 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 n, m;
int a[4086];
int son[4086][2];
int sk[4086], top;
long long f[4086][4086], g[4086];
int siz[4086];
void dfs(int i) {
  f[i][0] = f[i][1] = 0, siz[i] = 1;
  for (int j = 0; j <= 1; j++) {
    int to = son[i][j];
    if (!to) continue;
    dfs(to);
    for (int k = 0; k <= min(siz[i], m); k++) {
      for (int l = 0; l <= min(siz[to], m - k); l++) {
        g[k + l] = max(g[k + l],
                       f[i][k] + f[to][l] + 1ll * (a[to] - a[i]) * l * (m - l));
      }
    }
    for (int k = 0; k <= min(siz[i] + siz[to], m); k++)
      f[i][k] = g[k], g[k] = -1e18;
    siz[i] += siz[to];
  }
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    int last = 0;
    while (top && a[sk[top]] > a[i]) last = sk[top--];
    son[i][0] = last, son[sk[top]][1] = i;
    sk[++top] = i;
  }
  memset(f, -0x3f, sizeof(f)), memset(g, -0x3f, sizeof(g));
  dfs(sk[1]);
  printf("%lld", f[sk[1]][m]);
}

```