#include <bits/stdc++.h>
using namespace std;
int n, T, r, a[10000], cnt[301] = {0};
set<int> lis;
int conv(int v, int bi, int pi) { return v * (100 * 100) + bi * 100 + pi; }
void calc_lis() {
  lis.insert(a[0]);
  for (int i = (1); i < (n * r); i++) {
    auto it = lis.upper_bound(a[i]);
    if (it != lis.end()) lis.erase(it);
    lis.insert(a[i]);
  }
}
int main() {
  cin >> n >> T;
  r = min(n, T);
  for (int i = (0); i < (n); i++) {
    cin >> a[i];
    cnt[a[i]]++;
  }
  for (int i = (1); i < (r); i++)
    for (int j = (0); j < (n); j++) a[n * i + j] = a[j];
  for (int i = (0); i < (r); i++)
    for (int j = (0); j < (n); j++) a[n * i + j] = conv(a[n * i + j], i, j);
  calc_lis();
  int mx = -1;
  for (int i = (0); i < (301); i++) mx = max(mx, cnt[i]);
  cout << lis.size() + max(0, T - n) * mx;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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, r, a[10000], cnt[301] = {0};
set<int> lis;
int conv(int v, int bi, int pi) { return v * (100 * 100) + bi * 100 + pi; }
void calc_lis() {
  lis.insert(a[0]);
  for (int i = (1); i < (n * r); i++) {
    auto it = lis.upper_bound(a[i]);
    if (it != lis.end()) lis.erase(it);
    lis.insert(a[i]);
  }
}
int main() {
  cin >> n >> T;
  r = min(n, T);
  for (int i = (0); i < (n); i++) {
    cin >> a[i];
    cnt[a[i]]++;
  }
  for (int i = (1); i < (r); i++)
    for (int j = (0); j < (n); j++) a[n * i + j] = a[j];
  for (int i = (0); i < (r); i++)
    for (int j = (0); j < (n); j++) a[n * i + j] = conv(a[n * i + j], i, j);
  calc_lis();
  int mx = -1;
  for (int i = (0); i < (301); i++) mx = max(mx, cnt[i]);
  cout << lis.size() + max(0, T - n) * mx;
}

```