#include <bits/stdc++.h>
using namespace std;
const int N = 4010;
int n, m, a[N], rt, tot, L[N], siz[N], R[N], ls[N], rs[N];
long long val[N], dp[N][N];
inline void build(int &p, int l, int r) {
  if (l > r) return;
  p = ++tot;
  int pos = l;
  for (int i = l; i <= r; ++i)
    if (a[i] < a[pos]) pos = i;
  val[p] = a[pos];
  build(ls[p], l, pos - 1);
  build(rs[p], pos + 1, r);
  dp[p][0] = 0;
  dp[p][1] = m * val[p] - val[p];
  siz[p] = 1;
  for (int i = siz[p]; i >= 0; --i)
    for (int j = 0; j <= siz[ls[p]]; ++j)
      dp[p][i + j] =
          max(dp[p][i + j], dp[p][i] + dp[ls[p]][j] - 2ll * i * j * val[p]);
  siz[p] += siz[ls[p]];
  for (int i = siz[p]; i >= 0; --i)
    for (int j = 0; j <= siz[rs[p]]; ++j)
      dp[p][i + j] =
          max(dp[p][i + j], dp[p][i] + dp[rs[p]][j] - 2ll * i * j * val[p]);
  siz[p] += siz[rs[p]];
}
int main() {
  memset(dp, -0x3f, sizeof(dp));
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  build(rt, 1, n);
  printf("%lld\n", dp[1][m]);
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 = 4010;
int n, m, a[N], rt, tot, L[N], siz[N], R[N], ls[N], rs[N];
long long val[N], dp[N][N];
inline void build(int &p, int l, int r) {
  if (l > r) return;
  p = ++tot;
  int pos = l;
  for (int i = l; i <= r; ++i)
    if (a[i] < a[pos]) pos = i;
  val[p] = a[pos];
  build(ls[p], l, pos - 1);
  build(rs[p], pos + 1, r);
  dp[p][0] = 0;
  dp[p][1] = m * val[p] - val[p];
  siz[p] = 1;
  for (int i = siz[p]; i >= 0; --i)
    for (int j = 0; j <= siz[ls[p]]; ++j)
      dp[p][i + j] =
          max(dp[p][i + j], dp[p][i] + dp[ls[p]][j] - 2ll * i * j * val[p]);
  siz[p] += siz[ls[p]];
  for (int i = siz[p]; i >= 0; --i)
    for (int j = 0; j <= siz[rs[p]]; ++j)
      dp[p][i + j] =
          max(dp[p][i + j], dp[p][i] + dp[rs[p]][j] - 2ll * i * j * val[p]);
  siz[p] += siz[rs[p]];
}
int main() {
  memset(dp, -0x3f, sizeof(dp));
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  build(rt, 1, n);
  printf("%lld\n", dp[1][m]);
  return 0;
}

```