#include <bits/stdc++.h>
using namespace std;
const int N = 4e3 + 11;
int n, m;
long long a[N];
vector<pair<int, long long> > e[N];
void build(int l, int r, int fa) {
  if (l > r) return;
  int pos = l;
  for (int i = l + 1; i <= r; i++)
    if (a[i] < a[pos]) pos = i;
  e[fa].push_back(make_pair(pos, a[pos] - a[fa]));
  build(l, pos - 1, pos);
  build(pos + 1, r, pos);
}
int siz[N];
long long f[N][N], g[N];
void dfs(int w, int fa) {
  siz[w] = 1;
  for (pair<int, long long> v : e[w]) {
    if (v.first == fa) continue;
    dfs(v.first, w);
    for (int i = 0; i <= siz[w] + siz[v.first]; i++) g[i] = 0;
    for (int i = 0; i <= siz[w]; i++)
      for (int j = 0; j <= siz[v.first]; j++) {
        g[i + j] =
            max(g[i + j], f[w][i] + f[v.first][j] + j * (m - j) * v.second);
      }
    siz[w] += siz[v.first];
    for (int i = 0; i <= siz[w]; i++) f[w][i] = max(f[w][i], g[i]);
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> m;
  for (int i = 1; i <= n; i++) cin >> a[i];
  build(1, n, 0);
  dfs(e[0][0].first, 0);
  cout << f[e[0][0].first][m] << endl;
  return 0;
}

### 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;
const int N = 4e3 + 11;
int n, m;
long long a[N];
vector<pair<int, long long> > e[N];
void build(int l, int r, int fa) {
  if (l > r) return;
  int pos = l;
  for (int i = l + 1; i <= r; i++)
    if (a[i] < a[pos]) pos = i;
  e[fa].push_back(make_pair(pos, a[pos] - a[fa]));
  build(l, pos - 1, pos);
  build(pos + 1, r, pos);
}
int siz[N];
long long f[N][N], g[N];
void dfs(int w, int fa) {
  siz[w] = 1;
  for (pair<int, long long> v : e[w]) {
    if (v.first == fa) continue;
    dfs(v.first, w);
    for (int i = 0; i <= siz[w] + siz[v.first]; i++) g[i] = 0;
    for (int i = 0; i <= siz[w]; i++)
      for (int j = 0; j <= siz[v.first]; j++) {
        g[i + j] =
            max(g[i + j], f[w][i] + f[v.first][j] + j * (m - j) * v.second);
      }
    siz[w] += siz[v.first];
    for (int i = 0; i <= siz[w]; i++) f[w][i] = max(f[w][i], g[i]);
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> m;
  for (int i = 1; i <= n; i++) cin >> a[i];
  build(1, n, 0);
  dfs(e[0][0].first, 0);
  cout << f[e[0][0].first][m] << endl;
  return 0;
}

```