#include <bits/stdc++.h>
using namespace std;
const int MAXN = 4e3 + 5;
int m;
int a[MAXN];
vector<int> g[MAXN];
int siz[MAXN];
long long f[MAXN][MAXN];
void dfs_f(int u, int fa) {
  siz[u] = 1;
  f[u][0] = 0;
  f[u][1] = (long long)(m - 1) * a[u];
  for (int v : g[u])
    if (v != fa) {
      dfs_f(v, u);
      static long long nf[MAXN];
      memset(nf, 0xc0, (siz[u] + siz[v] + 1) << 3);
      for (int i = 0; i <= siz[u]; ++i)
        for (int j = 0; j <= siz[v]; ++j)
          nf[i + j] = max(nf[i + j], f[u][i] + f[v][j] - 2ll * i * j * a[u]);
      siz[u] += siz[v];
      memcpy(f[u], nf, (siz[u] + 1) << 3);
    }
}
int main(void) {
  int n;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  for (int i = 1; i <= n; ++i) g[i].resize(2);
  stack<int> sta;
  int rt = 0;
  for (int i = 1; i <= n; ++i) {
    while (sta.size() && a[sta.top()] > a[i]) g[i][0] = sta.top(), sta.pop();
    if (sta.size())
      g[sta.top()][1] = i;
    else
      rt = i;
    sta.push(i);
  }
  for (int i = 1; i <= n; ++i)
    for (int j = 0; j < (int)g[i].size(); ++j)
      if (!g[i][j]) {
        g[i].erase(g[i].begin() + j);
        --j;
      }
  dfs_f(rt, 0);
  printf("%lld", f[rt][m]);
  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 MAXN = 4e3 + 5;
int m;
int a[MAXN];
vector<int> g[MAXN];
int siz[MAXN];
long long f[MAXN][MAXN];
void dfs_f(int u, int fa) {
  siz[u] = 1;
  f[u][0] = 0;
  f[u][1] = (long long)(m - 1) * a[u];
  for (int v : g[u])
    if (v != fa) {
      dfs_f(v, u);
      static long long nf[MAXN];
      memset(nf, 0xc0, (siz[u] + siz[v] + 1) << 3);
      for (int i = 0; i <= siz[u]; ++i)
        for (int j = 0; j <= siz[v]; ++j)
          nf[i + j] = max(nf[i + j], f[u][i] + f[v][j] - 2ll * i * j * a[u]);
      siz[u] += siz[v];
      memcpy(f[u], nf, (siz[u] + 1) << 3);
    }
}
int main(void) {
  int n;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  for (int i = 1; i <= n; ++i) g[i].resize(2);
  stack<int> sta;
  int rt = 0;
  for (int i = 1; i <= n; ++i) {
    while (sta.size() && a[sta.top()] > a[i]) g[i][0] = sta.top(), sta.pop();
    if (sta.size())
      g[sta.top()][1] = i;
    else
      rt = i;
    sta.push(i);
  }
  for (int i = 1; i <= n; ++i)
    for (int j = 0; j < (int)g[i].size(); ++j)
      if (!g[i][j]) {
        g[i].erase(g[i].begin() + j);
        --j;
      }
  dfs_f(rt, 0);
  printf("%lld", f[rt][m]);
  return 0;
}

```