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 N = 4005;
int n, m, first[N];
int cnt, sta[N], c[N][2], siz[N];
long long f[N][N];
inline int read() {
  int x(0), y(1);
  char z(getchar());
  while (!isdigit(z)) {
    if (z == '-') y = -1;
    z = getchar();
  }
  while (isdigit(z)) x = x * 10 + (z ^ 48), z = getchar();
  return x * y;
}
inline void Max(long long &first, long long second) {
  first = first > second ? first : second;
}
inline void Min(long long &first, long long second) {
  first = first < second ? first : second;
}
inline int max(int &first, int &second) {
  return first > second ? first : second;
}
inline int min(int &first, int &second) {
  return first < second ? first : second;
}
inline void swap(int &first, int &second) {
  first ^= second, second ^= first, first ^= second;
}
inline int abs(int &first) { return first >= 0 ? first : -first; }
inline pair<int, int> operator+(const pair<int, int> &first,
                                const pair<int, int> &second) {
  return make_pair(first.first + second.first, first.second + second.second);
}
inline pair<int, int> operator-(const pair<int, int> &first,
                                const pair<int, int> &second) {
  return make_pair(first.first - second.first, first.second - second.second);
}
void Dfs(int x) {
  f[x][1] = 1ll * (m - 1) * first[x], siz[x] = 1;
  if (c[x][1]) {
    Dfs(c[x][1]), siz[x] += siz[c[x][1]];
    for (int i(min(siz[x], m)); i; --i)
      for (int j(min(siz[c[x][1]], i)); j; --j)
        Max(f[x][i],
            f[c[x][1]][j] + f[x][i - j] - 2ll * j * (i - j) * first[x]);
  }
  if (c[x][0]) {
    Dfs(c[x][0]), siz[x] += siz[c[x][0]];
    for (int i(min(siz[x], m)); i; --i)
      for (int j(min(siz[c[x][0]], i)); j; --j)
        Max(f[x][i],
            f[c[x][0]][j] + f[x][i - j] - 2ll * j * (i - j) * first[x]);
  }
}
signed main() {
  memset(f, -0x7f, sizeof f);
  n = read(), m = read();
  for (int i(1); i <= n; ++i) first[i] = read();
  for (int i(1); i <= n; ++i) {
    while (cnt && first[i] < first[sta[cnt]]) cnt--;
    c[i][1] = c[sta[cnt]][0], c[sta[cnt]][0] = i;
    sta[++cnt] = i;
  }
  Dfs(c[0][0]), printf("%lld\n", f[c[0][0]][m]);
}
