#include <bits/stdc++.h>
using namespace std;
inline long long read() {
  long long x = 0, f = 1, ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
inline void write(long long x) {
  if (x < 0) putchar('-'), x = -x;
  if (x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}
inline void writeln(long long x) {
  write(x);
  puts("");
}
const int N = 4005;
const long long INF = 1e15;
int n, m, mn[N][N], tot;
long long a[N], f[N][N];
short int mp[N][N];
bool g[N][N];
inline void init() {
  n = read();
  m = read();
  for (int i = 1; i <= n; i++) a[i] = read();
}
inline long long dp(int l, int r, int k) {
  if (k < 0) return -INF;
  if (l > r) {
    if (!k) return 0;
    return -INF;
  }
  if (l == r) {
    if (k == 0) return 0;
    if (k == 1) return 1ll * a[l] * (m - 1);
    return -INF;
  }
  if (!mp[l][r]) mp[l][r] = ++tot;
  int x = mp[l][r];
  if (g[x][k]) return f[x][k];
  g[x][k] = 1;
  int p = mn[l][r];
  if (p == l)
    return f[x][k] =
               max(dp(l + 1, r, k), dp(l + 1, r, k - 1) + a[l] * (m - 1ll) -
                                        2ll * a[l] * (k - 1));
  if (p == r)
    return f[x][k] =
               max(dp(l, r - 1, k), dp(l, r - 1, k - 1) + a[r] * (m - 1ll) -
                                        2ll * a[r] * (k - 1));
  if (p - l < r - p) {
    for (int i = 0; i <= min(k, p - l + 1); i++) {
      f[x][k] = max(f[x][k], dp(l, p, i) + dp(p + 1, r, k - i) -
                                 2ll * (k - i) * i * a[p]);
    }
  } else {
    for (int i = 0; i <= min(k, r - p + 1); i++) {
      f[x][k] = max(f[x][k], dp(l, p, k - i) + dp(p + 1, r, i) -
                                 2ll * (k - i) * i * a[p]);
    }
  }
  return f[x][k];
}
inline void solve() {
  for (int i = 1; i <= n; i++) {
    mn[i][i] = i;
    for (int j = i + 1; j <= n; j++) {
      if (a[j] < a[mn[i][j - 1]])
        mn[i][j] = j;
      else
        mn[i][j] = mn[i][j - 1];
    }
  }
  writeln(dp(1, n, m));
}
int main() {
  init();
  solve();
  return 0;
}

### Prompt

Create a solution in cpp for 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;
inline long long read() {
  long long x = 0, f = 1, ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
inline void write(long long x) {
  if (x < 0) putchar('-'), x = -x;
  if (x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}
inline void writeln(long long x) {
  write(x);
  puts("");
}
const int N = 4005;
const long long INF = 1e15;
int n, m, mn[N][N], tot;
long long a[N], f[N][N];
short int mp[N][N];
bool g[N][N];
inline void init() {
  n = read();
  m = read();
  for (int i = 1; i <= n; i++) a[i] = read();
}
inline long long dp(int l, int r, int k) {
  if (k < 0) return -INF;
  if (l > r) {
    if (!k) return 0;
    return -INF;
  }
  if (l == r) {
    if (k == 0) return 0;
    if (k == 1) return 1ll * a[l] * (m - 1);
    return -INF;
  }
  if (!mp[l][r]) mp[l][r] = ++tot;
  int x = mp[l][r];
  if (g[x][k]) return f[x][k];
  g[x][k] = 1;
  int p = mn[l][r];
  if (p == l)
    return f[x][k] =
               max(dp(l + 1, r, k), dp(l + 1, r, k - 1) + a[l] * (m - 1ll) -
                                        2ll * a[l] * (k - 1));
  if (p == r)
    return f[x][k] =
               max(dp(l, r - 1, k), dp(l, r - 1, k - 1) + a[r] * (m - 1ll) -
                                        2ll * a[r] * (k - 1));
  if (p - l < r - p) {
    for (int i = 0; i <= min(k, p - l + 1); i++) {
      f[x][k] = max(f[x][k], dp(l, p, i) + dp(p + 1, r, k - i) -
                                 2ll * (k - i) * i * a[p]);
    }
  } else {
    for (int i = 0; i <= min(k, r - p + 1); i++) {
      f[x][k] = max(f[x][k], dp(l, p, k - i) + dp(p + 1, r, i) -
                                 2ll * (k - i) * i * a[p]);
    }
  }
  return f[x][k];
}
inline void solve() {
  for (int i = 1; i <= n; i++) {
    mn[i][i] = i;
    for (int j = i + 1; j <= n; j++) {
      if (a[j] < a[mn[i][j - 1]])
        mn[i][j] = j;
      else
        mn[i][j] = mn[i][j - 1];
    }
  }
  writeln(dp(1, n, m));
}
int main() {
  init();
  solve();
  return 0;
}

```