#include <bits/stdc++.h>
using namespace std;
const int N = 4002;
long long g[N][N];
int st[N], c[N][2], f[N], siz[N], a[N];
int n, m, i, tp, rt;
void dfs(int x) {
  int lc = c[x][0], rc = c[x][1];
  siz[x] = 1;
  if (!lc && !rc) {
    g[x][0] = 0;
    g[x][1] = (long long)a[x] * (m - 1);
    return;
  }
  if (!lc && rc) swap(lc, rc);
  if (!rc) {
    dfs(lc);
    siz[x] += siz[lc];
    memcpy(g[x], g[lc], (siz[lc] + 1) * sizeof g[0][0]);
    for (int i = 0; i <= siz[lc]; i++)
      g[x][i + 1] = max(g[x][i + 1],
                        g[lc][i] + (long long)(m - 1) * a[x] - 2ll * i * a[x]);
    return;
  }
  dfs(lc);
  dfs(rc);
  siz[x] += siz[lc] + siz[rc];
  for (int i = 0; i <= siz[lc]; i++)
    for (int j = 0; j <= siz[rc]; j++)
      g[x][i + j] = max(g[x][i + j], g[lc][i] + g[rc][j] - 2ll * i * j * a[x]);
  for (int i = 0; i <= siz[lc]; i++)
    for (int j = 0; j <= siz[rc]; j++)
      g[x][i + j + 1] =
          max(g[x][i + j + 1], g[lc][i] + g[rc][j] + (long long)(m - 1) * a[x] -
                                   2ll * (i * j + i + j) * a[x]);
}
int main() {
  memset(g, -0x3f, sizeof g);
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m;
  for (i = 1; i <= n; i++) {
    cin >> a[i];
    st[tp + 1] = 0;
    while (tp && a[st[tp]] > a[i]) --tp;
    c[st[tp]][1] = i;
    c[i][0] = st[tp + 1];
    st[++tp] = i;
  }
  for (i = 1; i <= n; i++) f[c[i][0]] = f[c[i][1]] = i;
  rt = min_element(f + 1, f + n + 1) - f;
  ;
  dfs(rt);
  cout << g[rt][m] << endl;
}

### Prompt

Your challenge is to write 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 = 4002;
long long g[N][N];
int st[N], c[N][2], f[N], siz[N], a[N];
int n, m, i, tp, rt;
void dfs(int x) {
  int lc = c[x][0], rc = c[x][1];
  siz[x] = 1;
  if (!lc && !rc) {
    g[x][0] = 0;
    g[x][1] = (long long)a[x] * (m - 1);
    return;
  }
  if (!lc && rc) swap(lc, rc);
  if (!rc) {
    dfs(lc);
    siz[x] += siz[lc];
    memcpy(g[x], g[lc], (siz[lc] + 1) * sizeof g[0][0]);
    for (int i = 0; i <= siz[lc]; i++)
      g[x][i + 1] = max(g[x][i + 1],
                        g[lc][i] + (long long)(m - 1) * a[x] - 2ll * i * a[x]);
    return;
  }
  dfs(lc);
  dfs(rc);
  siz[x] += siz[lc] + siz[rc];
  for (int i = 0; i <= siz[lc]; i++)
    for (int j = 0; j <= siz[rc]; j++)
      g[x][i + j] = max(g[x][i + j], g[lc][i] + g[rc][j] - 2ll * i * j * a[x]);
  for (int i = 0; i <= siz[lc]; i++)
    for (int j = 0; j <= siz[rc]; j++)
      g[x][i + j + 1] =
          max(g[x][i + j + 1], g[lc][i] + g[rc][j] + (long long)(m - 1) * a[x] -
                                   2ll * (i * j + i + j) * a[x]);
}
int main() {
  memset(g, -0x3f, sizeof g);
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m;
  for (i = 1; i <= n; i++) {
    cin >> a[i];
    st[tp + 1] = 0;
    while (tp && a[st[tp]] > a[i]) --tp;
    c[st[tp]][1] = i;
    c[i][0] = st[tp + 1];
    st[++tp] = i;
  }
  for (i = 1; i <= n; i++) f[c[i][0]] = f[c[i][1]] = i;
  rt = min_element(f + 1, f + n + 1) - f;
  ;
  dfs(rt);
  cout << g[rt][m] << endl;
}

```