#include <bits/stdc++.h>
using namespace std;
int ls[4010], rs[4010];
int lw[4010], rw[4010];
int n, a[4010], st[4010], cnt, m;
long long f[4010][4010];
int sz[4010];
void dfs(int x) {
  sz[x]++;
  if (ls[x]) dfs(ls[x]), sz[x] += sz[ls[x]];
  if (rs[x]) dfs(rs[x]), sz[x] += sz[rs[x]];
  for (int i = 0; i <= sz[ls[x]]; i++)
    for (int j = 0; j <= sz[rs[x]]; j++) {
      if (i + j > m)
        break;
      else {
        f[x][i + j] =
            max(f[ls[x]][i] + f[rs[x]][j] + 1ll * i * (m - i) * lw[x] +
                    1ll * j * (m - j) * rw[x],
                f[x][i + j]);
        f[x][i + j + 1] =
            max(f[ls[x]][i] + f[rs[x]][j] + 1ll * i * (m - i) * lw[x] +
                    1ll * j * (m - j) * rw[x],
                f[x][i + j + 1]);
      }
    }
}
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++) {
    int pre = cnt;
    while (cnt && a[i] < a[st[cnt]]) cnt--;
    if (cnt) {
      rs[st[cnt]] = i;
      rw[st[cnt]] = a[i] - a[st[cnt]];
    }
    if (cnt < pre) {
      ls[i] = st[cnt + 1];
      lw[i] = a[st[cnt + 1]] - a[i];
    }
    st[++cnt] = i;
  }
  dfs(st[1]);
  printf("%lld\n", f[st[1]][m]);
}

### 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;
int ls[4010], rs[4010];
int lw[4010], rw[4010];
int n, a[4010], st[4010], cnt, m;
long long f[4010][4010];
int sz[4010];
void dfs(int x) {
  sz[x]++;
  if (ls[x]) dfs(ls[x]), sz[x] += sz[ls[x]];
  if (rs[x]) dfs(rs[x]), sz[x] += sz[rs[x]];
  for (int i = 0; i <= sz[ls[x]]; i++)
    for (int j = 0; j <= sz[rs[x]]; j++) {
      if (i + j > m)
        break;
      else {
        f[x][i + j] =
            max(f[ls[x]][i] + f[rs[x]][j] + 1ll * i * (m - i) * lw[x] +
                    1ll * j * (m - j) * rw[x],
                f[x][i + j]);
        f[x][i + j + 1] =
            max(f[ls[x]][i] + f[rs[x]][j] + 1ll * i * (m - i) * lw[x] +
                    1ll * j * (m - j) * rw[x],
                f[x][i + j + 1]);
      }
    }
}
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++) {
    int pre = cnt;
    while (cnt && a[i] < a[st[cnt]]) cnt--;
    if (cnt) {
      rs[st[cnt]] = i;
      rw[st[cnt]] = a[i] - a[st[cnt]];
    }
    if (cnt < pre) {
      ls[i] = st[cnt + 1];
      lw[i] = a[st[cnt + 1]] - a[i];
    }
    st[++cnt] = i;
  }
  dfs(st[1]);
  printf("%lld\n", f[st[1]][m]);
}

```