#include <bits/stdc++.h>
using namespace std;
const long long N = 4e3 + 1e2 + 7;
long long n, m, st[N], top, a[N];
long long f[N][N], tot, g[N];
long long dfs(long long l, long long r) {
  long long p = min_element(a + l, a + r + 1) - a;
  long long x = ++tot;
  for (long long j = 0; j <= r - l + 1; j++) f[x][j] = -1e18;
  f[x][0] = 0;
  f[x][1] = a[p] * (m - 1);
  long long ls = -1, rs = -1;
  if (l <= p - 1) ls = dfs(l, p - 1);
  if (p + 1 <= r) rs = dfs(p + 1, r);
  long long sz = 1;
  if (ls != -1) {
    long long szl = p - l;
    for (long long j = 0; j <= szl + sz; j++) g[j] = -1e18;
    for (long long j = 0; j <= szl; j++)
      for (long long k = 0; k <= sz; k++)
        g[j + k] = max(g[j + k], f[ls][j] + f[x][k] - 2 * j * k * a[p]);
    sz += szl;
    for (long long j = 0; j <= sz; j++) f[x][j] = max(f[x][j], g[j]);
  }
  if (rs != -1) {
    long long szr = r - p;
    for (long long j = 0; j <= szr + sz; j++) g[j] = -1e18;
    for (long long j = 0; j <= szr; j++)
      for (long long k = 0; k <= sz; k++)
        g[j + k] = max(g[j + k], f[rs][j] + f[x][k] - 2 * j * k * a[p]);
    sz += szr;
    for (long long j = 0; j <= sz; j++) f[x][j] = max(f[x][j], g[j]);
  }
  return x;
}
signed main() {
  scanf("%lld%lld", &n, &m);
  for (long long i = 1; i <= n; i++) scanf("%lld", &a[i]);
  dfs(1, n);
  printf("%lld\n", f[1][m]);
}

### Prompt

Please create a solution in cpp 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 long long N = 4e3 + 1e2 + 7;
long long n, m, st[N], top, a[N];
long long f[N][N], tot, g[N];
long long dfs(long long l, long long r) {
  long long p = min_element(a + l, a + r + 1) - a;
  long long x = ++tot;
  for (long long j = 0; j <= r - l + 1; j++) f[x][j] = -1e18;
  f[x][0] = 0;
  f[x][1] = a[p] * (m - 1);
  long long ls = -1, rs = -1;
  if (l <= p - 1) ls = dfs(l, p - 1);
  if (p + 1 <= r) rs = dfs(p + 1, r);
  long long sz = 1;
  if (ls != -1) {
    long long szl = p - l;
    for (long long j = 0; j <= szl + sz; j++) g[j] = -1e18;
    for (long long j = 0; j <= szl; j++)
      for (long long k = 0; k <= sz; k++)
        g[j + k] = max(g[j + k], f[ls][j] + f[x][k] - 2 * j * k * a[p]);
    sz += szl;
    for (long long j = 0; j <= sz; j++) f[x][j] = max(f[x][j], g[j]);
  }
  if (rs != -1) {
    long long szr = r - p;
    for (long long j = 0; j <= szr + sz; j++) g[j] = -1e18;
    for (long long j = 0; j <= szr; j++)
      for (long long k = 0; k <= sz; k++)
        g[j + k] = max(g[j + k], f[rs][j] + f[x][k] - 2 * j * k * a[p]);
    sz += szr;
    for (long long j = 0; j <= sz; j++) f[x][j] = max(f[x][j], g[j]);
  }
  return x;
}
signed main() {
  scanf("%lld%lld", &n, &m);
  for (long long i = 1; i <= n; i++) scanf("%lld", &a[i]);
  dfs(1, n);
  printf("%lld\n", f[1][m]);
}

```