#include <bits/stdc++.h>
using namespace std;
int a[101];
int b[301];
void exec() {
  int n;
  long long T;
  cin >> n >> T;
  for (int i = 0; i < n; i++) cin >> a[i], b[a[i]]++;
  int mx = max_element(b, b + 301) - b;
  int len;
  if (T < 2 * n)
    len = n * T;
  else
    len = 2 * n * n;
  vector<int> v;
  v.reserve(len);
  int num;
  int i = 0;
  while (i < len) {
    num = a[i % n];
    if (v.size() == 0 || num >= v.back())
      v.push_back(num);
    else
      *upper_bound(v.begin(), v.end(), num) = num;
    i++;
  }
  long long x = T - (len / n);
  cerr << "\n" << mx << "\n";
  for (auto i : v) cerr << i << " ";
  cerr << "\n";
  cout << v.size() + x * count(a, a + n, mx);
}
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  exec();
  return 0;
}

### Prompt

Generate a cpp solution to the following problem:

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 a[101];
int b[301];
void exec() {
  int n;
  long long T;
  cin >> n >> T;
  for (int i = 0; i < n; i++) cin >> a[i], b[a[i]]++;
  int mx = max_element(b, b + 301) - b;
  int len;
  if (T < 2 * n)
    len = n * T;
  else
    len = 2 * n * n;
  vector<int> v;
  v.reserve(len);
  int num;
  int i = 0;
  while (i < len) {
    num = a[i % n];
    if (v.size() == 0 || num >= v.back())
      v.push_back(num);
    else
      *upper_bound(v.begin(), v.end(), num) = num;
    i++;
  }
  long long x = T - (len / n);
  cerr << "\n" << mx << "\n";
  for (auto i : v) cerr << i << " ";
  cerr << "\n";
  cout << v.size() + x * count(a, a + n, mx);
}
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  exec();
  return 0;
}

```