#include <bits/stdc++.h>
using namespace std;
int n, m, a[4010], ar[4010], sz, lc[4010], rc[4010], siz[4010];
long long dp[4010][4010];
void DP(const int& x, const int& fa) {
  if (!x) return;
  DP(lc[x], x), DP(rc[x], x), siz[x] = siz[lc[x]] + siz[rc[x]] + 1;
  for (int i = 0; i <= siz[lc[x]]; ++i)
    for (int j = 0; j <= siz[rc[x]] && i + j <= m; ++j) {
      dp[x][i + j] =
          max(dp[x][i + j], dp[lc[x]][i] + dp[rc[x]][j] +
                                1ll * (a[x] - a[fa]) * (m - i - j) * (i + j));
      dp[x][i + j + 1] =
          max(dp[x][i + j + 1],
              dp[lc[x]][i] + dp[rc[x]][j] +
                  1ll * (a[x] - a[fa]) * (m - i - j - 1) * (i + j + 1));
    }
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  for (int i = 1; i <= n; ++i) {
    while (sz && a[ar[sz]] > a[i]) lc[i] = ar[sz--];
    rc[ar[sz]] = i, ar[++sz] = i;
  }
  for (int i = sz; i > 1; --i) rc[ar[i - 1]] = ar[i];
  a[0] = a[ar[1]], DP(ar[1], 0), printf("%lld", dp[ar[1]][m]);
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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;
int n, m, a[4010], ar[4010], sz, lc[4010], rc[4010], siz[4010];
long long dp[4010][4010];
void DP(const int& x, const int& fa) {
  if (!x) return;
  DP(lc[x], x), DP(rc[x], x), siz[x] = siz[lc[x]] + siz[rc[x]] + 1;
  for (int i = 0; i <= siz[lc[x]]; ++i)
    for (int j = 0; j <= siz[rc[x]] && i + j <= m; ++j) {
      dp[x][i + j] =
          max(dp[x][i + j], dp[lc[x]][i] + dp[rc[x]][j] +
                                1ll * (a[x] - a[fa]) * (m - i - j) * (i + j));
      dp[x][i + j + 1] =
          max(dp[x][i + j + 1],
              dp[lc[x]][i] + dp[rc[x]][j] +
                  1ll * (a[x] - a[fa]) * (m - i - j - 1) * (i + j + 1));
    }
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  for (int i = 1; i <= n; ++i) {
    while (sz && a[ar[sz]] > a[i]) lc[i] = ar[sz--];
    rc[ar[sz]] = i, ar[++sz] = i;
  }
  for (int i = sz; i > 1; --i) rc[ar[i - 1]] = ar[i];
  a[0] = a[ar[1]], DP(ar[1], 0), printf("%lld", dp[ar[1]][m]);
  return 0;
}

```