#include <bits/stdc++.h>
using namespace std;
const int maxn = 4005;
int n, m, u, tot = 0;
int siz[maxn], a[maxn], b[maxn];
int ch[maxn][2];
long long f[maxn][maxn];
inline int build(int l, int r, int sub) {
  if (l > r) return 0;
  int u = ++tot;
  int ptr = -1;
  for (int i = l; i <= r; ++i)
    if (!~ptr || a[i] < a[ptr]) ptr = i;
  siz[u] = 1;
  int p, q;
  p = ch[u][0] = build(l, ptr - 1, a[ptr]);
  siz[u] += siz[p];
  q = ch[u][1] = build(ptr + 1, r, a[ptr]);
  siz[u] += siz[q];
  a[ptr] -= sub, b[u] = a[ptr];
  for (int i = 0; i <= siz[u]; ++i) {
    for (int j = max(0, i - 1 - siz[q]); j <= min(i - 1, siz[p]); ++j) {
      int k = i - 1 - j;
      long long res = f[p][j] + f[q][k];
      f[u][i] = max(f[u][i], res);
    }
    for (int j = max(0, i - siz[q]); j <= min(i, siz[p]); ++j) {
      int k = i - j;
      long long res = f[p][j] + f[q][k];
      f[u][i] = max(f[u][i], res);
    }
    if (u) f[u][i] += 1ll * i * (m - i) * a[ptr];
  }
  return u;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 0; i < n; ++i) scanf("%d", a + i);
  int rt = build(0, n - 1, 0);
  printf("%lld\n", f[rt][m]);
  return 0;
}

### Prompt

Please formulate 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 = 4005;
int n, m, u, tot = 0;
int siz[maxn], a[maxn], b[maxn];
int ch[maxn][2];
long long f[maxn][maxn];
inline int build(int l, int r, int sub) {
  if (l > r) return 0;
  int u = ++tot;
  int ptr = -1;
  for (int i = l; i <= r; ++i)
    if (!~ptr || a[i] < a[ptr]) ptr = i;
  siz[u] = 1;
  int p, q;
  p = ch[u][0] = build(l, ptr - 1, a[ptr]);
  siz[u] += siz[p];
  q = ch[u][1] = build(ptr + 1, r, a[ptr]);
  siz[u] += siz[q];
  a[ptr] -= sub, b[u] = a[ptr];
  for (int i = 0; i <= siz[u]; ++i) {
    for (int j = max(0, i - 1 - siz[q]); j <= min(i - 1, siz[p]); ++j) {
      int k = i - 1 - j;
      long long res = f[p][j] + f[q][k];
      f[u][i] = max(f[u][i], res);
    }
    for (int j = max(0, i - siz[q]); j <= min(i, siz[p]); ++j) {
      int k = i - j;
      long long res = f[p][j] + f[q][k];
      f[u][i] = max(f[u][i], res);
    }
    if (u) f[u][i] += 1ll * i * (m - i) * a[ptr];
  }
  return u;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 0; i < n; ++i) scanf("%d", a + i);
  int rt = build(0, n - 1, 0);
  printf("%lld\n", f[rt][m]);
  return 0;
}

```