#include <bits/stdc++.h>
using namespace std;
const long long N = 4e3 + 5;
long long n, m, top, a[N], f[N][N], ls[N], rs[N], stk[N], siz[N];
void dfs(long long x) {
  siz[x] = 1;
  if (ls[x]) dfs(ls[x]), siz[x] += siz[ls[x]];
  if (rs[x]) dfs(rs[x]), siz[x] += siz[rs[x]];
}
void solve(long long p) {
  if (!p) return;
  solve(ls[p]), solve(rs[p]);
  memset(f[p], -0x3f, sizeof(f[p]));
  for (long long i = 0; i <= siz[ls[p]]; i++)
    for (long long j = 0; j <= siz[rs[p]]; j++) {
      f[p][i + j] =
          max(f[p][i + j], f[ls[p]][i] + f[rs[p]][j] - 2 * i * j * a[p]);
      f[p][i + j + 1] =
          max(f[p][i + j + 1], f[ls[p]][i] + f[rs[p]][j] - 2 * i * j * a[p] -
                                   2 * (i + j) * a[p] + (m - 1) * a[p]);
    }
}
signed main() {
  cin >> n >> m;
  a[0] = 1e18;
  long long p = 0;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
    if (a[i] < a[p]) p = i;
  }
  for (long long i = 1; i <= n; i++) {
    long long k = top;
    while (k && a[stk[k]] > a[i]) k--;
    if (k) rs[stk[k]] = i;
    if (k < top) ls[i] = stk[k + 1];
    stk[++k] = i;
    top = k;
  }
  dfs(p);
  solve(p);
  cout << f[p][m] << endl;
}

### 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;
const long long N = 4e3 + 5;
long long n, m, top, a[N], f[N][N], ls[N], rs[N], stk[N], siz[N];
void dfs(long long x) {
  siz[x] = 1;
  if (ls[x]) dfs(ls[x]), siz[x] += siz[ls[x]];
  if (rs[x]) dfs(rs[x]), siz[x] += siz[rs[x]];
}
void solve(long long p) {
  if (!p) return;
  solve(ls[p]), solve(rs[p]);
  memset(f[p], -0x3f, sizeof(f[p]));
  for (long long i = 0; i <= siz[ls[p]]; i++)
    for (long long j = 0; j <= siz[rs[p]]; j++) {
      f[p][i + j] =
          max(f[p][i + j], f[ls[p]][i] + f[rs[p]][j] - 2 * i * j * a[p]);
      f[p][i + j + 1] =
          max(f[p][i + j + 1], f[ls[p]][i] + f[rs[p]][j] - 2 * i * j * a[p] -
                                   2 * (i + j) * a[p] + (m - 1) * a[p]);
    }
}
signed main() {
  cin >> n >> m;
  a[0] = 1e18;
  long long p = 0;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
    if (a[i] < a[p]) p = i;
  }
  for (long long i = 1; i <= n; i++) {
    long long k = top;
    while (k && a[stk[k]] > a[i]) k--;
    if (k) rs[stk[k]] = i;
    if (k < top) ls[i] = stk[k + 1];
    stk[++k] = i;
    top = k;
  }
  dfs(p);
  solve(p);
  cout << f[p][m] << endl;
}

```