#include <bits/stdc++.h>
using namespace std;
int n, t, a[10005], f[10005], num[305];
int main() {
  cin >> n >> t;
  int maxlen = min(n * n, n * t);
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    num[a[i]]++;
  }
  for (int i = n + 1; i <= maxlen; i++) a[i] = a[i - n];
  f[1] = 1;
  for (int i = 2; i <= maxlen; i++) {
    for (int j = 1; j < i; j++)
      if (a[j] <= a[i]) f[i] = max(f[i], f[j]);
    f[i]++;
  }
  int ans = 0, maxnum = 0;
  for (int i = 1; i <= maxlen; i++) ans = max(ans, f[i]);
  for (int i = 1; i <= 300; i++) maxnum = max(maxnum, num[i]);
  if (n > t)
    cout << ans << endl;
  else
    cout << ans + (t - n) * maxnum << endl;
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

You are given an array of positive integers a1, a2, ..., an × T of length n × T. We know that for any i > n it is true that ai = ai - n. Find the length of the longest non-decreasing sequence of the given array.

Input

The first line contains two space-separated integers: n, T (1 ≤ n ≤ 100, 1 ≤ T ≤ 107). The second line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 300).

Output

Print a single number — the length of a sought sequence.

Examples

Input

4 3
3 1 4 2


Output

5

Note

The array given in the sample looks like that: 3, 1, 4, 2, 3, 1, 4, 2, 3, 1, 4, 2. The elements in bold form the largest non-decreasing subsequence. 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int n, t, a[10005], f[10005], num[305];
int main() {
  cin >> n >> t;
  int maxlen = min(n * n, n * t);
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    num[a[i]]++;
  }
  for (int i = n + 1; i <= maxlen; i++) a[i] = a[i - n];
  f[1] = 1;
  for (int i = 2; i <= maxlen; i++) {
    for (int j = 1; j < i; j++)
      if (a[j] <= a[i]) f[i] = max(f[i], f[j]);
    f[i]++;
  }
  int ans = 0, maxnum = 0;
  for (int i = 1; i <= maxlen; i++) ans = max(ans, f[i]);
  for (int i = 1; i <= 300; i++) maxnum = max(maxnum, num[i]);
  if (n > t)
    cout << ans << endl;
  else
    cout << ans + (t - n) * maxnum << endl;
  return 0;
}

```