#include <bits/stdc++.h>
using namespace std;
inline void read(int &x) {
  char c;
  int f = 1;
  while (!isdigit(c = getchar()))
    if (c == '-') f = -1;
  x = (c & 15);
  while (isdigit(c = getchar())) x = (x << 1) + (x << 3) + (c & 15);
  x *= f;
}
int n, m, i, j, a[4005], sz[4005];
int stk[4005], tp, lc[4005], rc[4005];
long long f[4005][4005], g[4005];
vector<int> bi[4005];
void dfs(int x, int fa) {
  memset(f[x], -0x3f, sizeof(f[x]));
  f[x][0] = 0;
  f[x][1] = 1ll * (m - 1) * a[x];
  sz[x] = 1;
  for (__typeof((bi[x]).begin()) it = (bi[x]).begin(); it != (bi[x]).end();
       ++it) {
    dfs(*it, x);
    memset(g, -0x3f, sizeof(g));
    int i, j;
    for ((i) = 0; (i) <= (sz[x]); (i)++)
      for ((j) = 0; (j) <= (sz[*it]); (j)++) {
        g[i + j] = max(g[i + j], f[x][i] + f[*it][j] - 2ll * i * j * a[x]);
      }
    memcpy(f[x], g, sizeof(f[x]));
    sz[x] += sz[*it];
  }
}
int main() {
  read(n);
  read(m);
  for ((i) = 1; (i) <= (n); (i)++) read(a[i]);
  for ((i) = 1; (i) <= (n); (i)++) {
    int lst = 0;
    while (tp && a[stk[tp]] >= a[i]) {
      lst = stk[tp--];
    }
    lc[i] = lst;
    if (tp) rc[stk[tp]] = i;
    stk[++tp] = i;
  }
  int rt = stk[1];
  for ((i) = 1; (i) <= (n); (i)++) {
    if (lc[i]) bi[i].push_back(lc[i]);
    if (rc[i]) bi[i].push_back(rc[i]);
  }
  dfs(rt, 0);
  cout << f[rt][m] << endl;
  return 0;
}

### 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;
inline void read(int &x) {
  char c;
  int f = 1;
  while (!isdigit(c = getchar()))
    if (c == '-') f = -1;
  x = (c & 15);
  while (isdigit(c = getchar())) x = (x << 1) + (x << 3) + (c & 15);
  x *= f;
}
int n, m, i, j, a[4005], sz[4005];
int stk[4005], tp, lc[4005], rc[4005];
long long f[4005][4005], g[4005];
vector<int> bi[4005];
void dfs(int x, int fa) {
  memset(f[x], -0x3f, sizeof(f[x]));
  f[x][0] = 0;
  f[x][1] = 1ll * (m - 1) * a[x];
  sz[x] = 1;
  for (__typeof((bi[x]).begin()) it = (bi[x]).begin(); it != (bi[x]).end();
       ++it) {
    dfs(*it, x);
    memset(g, -0x3f, sizeof(g));
    int i, j;
    for ((i) = 0; (i) <= (sz[x]); (i)++)
      for ((j) = 0; (j) <= (sz[*it]); (j)++) {
        g[i + j] = max(g[i + j], f[x][i] + f[*it][j] - 2ll * i * j * a[x]);
      }
    memcpy(f[x], g, sizeof(f[x]));
    sz[x] += sz[*it];
  }
}
int main() {
  read(n);
  read(m);
  for ((i) = 1; (i) <= (n); (i)++) read(a[i]);
  for ((i) = 1; (i) <= (n); (i)++) {
    int lst = 0;
    while (tp && a[stk[tp]] >= a[i]) {
      lst = stk[tp--];
    }
    lc[i] = lst;
    if (tp) rc[stk[tp]] = i;
    stk[++tp] = i;
  }
  int rt = stk[1];
  for ((i) = 1; (i) <= (n); (i)++) {
    if (lc[i]) bi[i].push_back(lc[i]);
    if (rc[i]) bi[i].push_back(rc[i]);
  }
  dfs(rt, 0);
  cout << f[rt][m] << endl;
  return 0;
}

```