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].
#include <bits/stdc++.h>
using namespace std;
const int MX = 4005;
int n, m, a[MX];
vector<long long> f(int i, int j) {
  if (i > j) return {0};
  int ind = i;
  for (int k = i; k <= j; k++)
    if (a[k] < a[ind]) ind = k;
  vector<long long> izq = f(i, ind - 1);
  vector<long long> der = f(ind + 1, j);
  vector<long long> res(j - i + 2, -1e18);
  for (int x = 0; x < int(izq.size()); x++)
    for (int y = 0; y < int(der.size()); y++) {
      res[x + y] = max(res[x + y], izq[x] + der[y] - 2ll * x * y * a[ind]);
      res[x + y + 1] =
          max(res[x + y + 1], izq[x] + der[y] - 2ll * (x * y + x + y) * a[ind] +
                                  (m - 1) * a[ind]);
    }
  return res;
}
void main_() {
  cin >> n >> m;
  for (int i = 0; i < int(n); i++) cin >> a[i];
  cout << f(0, n - 1)[m] << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) main_();
  return 0;
}
