#include <bits/stdc++.h>
const int N = 4086;
int n, m, rt, tp, st[N], sze[N], ls[N], rs[N];
long long f[N][N], a[N];
long long Max(long long a, long long b) { return a > b ? a : b; }
int Min(int a, int b) { return a < b ? a : b; }
int Abs(int x) { return x < 0 ? -x : x; }
long long read() {
  char ch = getchar();
  long long nn = 0, ssss = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') ssss *= -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    nn = nn * 10 + (ch - '0');
    ch = getchar();
  }
  return nn * ssss;
}
bool dfs(int np) {
  int lw = Abs(a[ls[np]] - a[np]), rw = Abs(a[rs[np]] - a[np]);
  sze[np] = 1;
  if (ls[np]) {
    dfs(ls[np]);
    for (int i = Min(sze[np], m); i + 1; i--)
      for (int j = Min(sze[ls[np]], m); j + 1; j--)
        f[np][i + j] =
            Max(f[np][i + j], f[np][i] + f[ls[np]][j] + 1ll * j * (m - j) * lw);
    sze[np] += sze[ls[np]];
  }
  if (rs[np]) {
    dfs(rs[np]);
    for (int i = Min(sze[np], m); i + 1; i--)
      for (int j = Min(sze[rs[np]], m); j + 1; j--)
        f[np][i + j] =
            Max(f[np][i + j], f[np][i] + f[rs[np]][j] + 1ll * j * (m - j) * rw);
    sze[np] += sze[rs[np]];
  }
  return true;
}
int main() {
  n = (int)read();
  m = (int)read();
  a[rt = n + 1] = 1145141919000;
  for (int i = 1; i <= n; i++) {
    a[i] = read();
    int lt = tp;
    while (tp && a[st[tp]] > a[i]) tp--;
    if (tp) rs[st[tp]] = i;
    if (tp < lt) ls[i] = st[tp + 1];
    st[++tp] = i;
    if (a[i] < a[rt]) rt = i;
  }
  dfs(rt);
  printf("%lld", f[rt][m]);
}

### 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>
const int N = 4086;
int n, m, rt, tp, st[N], sze[N], ls[N], rs[N];
long long f[N][N], a[N];
long long Max(long long a, long long b) { return a > b ? a : b; }
int Min(int a, int b) { return a < b ? a : b; }
int Abs(int x) { return x < 0 ? -x : x; }
long long read() {
  char ch = getchar();
  long long nn = 0, ssss = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') ssss *= -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    nn = nn * 10 + (ch - '0');
    ch = getchar();
  }
  return nn * ssss;
}
bool dfs(int np) {
  int lw = Abs(a[ls[np]] - a[np]), rw = Abs(a[rs[np]] - a[np]);
  sze[np] = 1;
  if (ls[np]) {
    dfs(ls[np]);
    for (int i = Min(sze[np], m); i + 1; i--)
      for (int j = Min(sze[ls[np]], m); j + 1; j--)
        f[np][i + j] =
            Max(f[np][i + j], f[np][i] + f[ls[np]][j] + 1ll * j * (m - j) * lw);
    sze[np] += sze[ls[np]];
  }
  if (rs[np]) {
    dfs(rs[np]);
    for (int i = Min(sze[np], m); i + 1; i--)
      for (int j = Min(sze[rs[np]], m); j + 1; j--)
        f[np][i + j] =
            Max(f[np][i + j], f[np][i] + f[rs[np]][j] + 1ll * j * (m - j) * rw);
    sze[np] += sze[rs[np]];
  }
  return true;
}
int main() {
  n = (int)read();
  m = (int)read();
  a[rt = n + 1] = 1145141919000;
  for (int i = 1; i <= n; i++) {
    a[i] = read();
    int lt = tp;
    while (tp && a[st[tp]] > a[i]) tp--;
    if (tp) rs[st[tp]] = i;
    if (tp < lt) ls[i] = st[tp + 1];
    st[++tp] = i;
    if (a[i] < a[rt]) rt = i;
  }
  dfs(rt);
  printf("%lld", f[rt][m]);
}

```