#include <bits/stdc++.h>
using namespace std;
const int M = 105;
const int D = 305;
int a[M];
vector<int> v;
int cnt[D];
int LIS() {
  multiset<int> dp;
  dp.insert(-1);
  for (auto x : v) {
    if (x < *dp.rbegin()) dp.erase(dp.upper_bound(x));
    dp.insert(x);
  }
  return dp.size() - 1;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  ;
  int n, t;
  cin >> n >> t;
  int x = min(2 * n, t);
  int MX = 0;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    cnt[a[i]]++;
    ((MX) = max((MX), (cnt[a[i]])));
  }
  for (int k = 0; k < x; k++)
    for (int i = 0; i < n; i++) v.push_back(a[i]);
  cout << LIS() + (t - x) * MX << endl;
  return 0;
}

### Prompt

Please create a solution in CPP 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;
const int M = 105;
const int D = 305;
int a[M];
vector<int> v;
int cnt[D];
int LIS() {
  multiset<int> dp;
  dp.insert(-1);
  for (auto x : v) {
    if (x < *dp.rbegin()) dp.erase(dp.upper_bound(x));
    dp.insert(x);
  }
  return dp.size() - 1;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  ;
  int n, t;
  cin >> n >> t;
  int x = min(2 * n, t);
  int MX = 0;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    cnt[a[i]]++;
    ((MX) = max((MX), (cnt[a[i]])));
  }
  for (int k = 0; k < x; k++)
    for (int i = 0; i < n; i++) v.push_back(a[i]);
  cout << LIS() + (t - x) * MX << endl;
  return 0;
}

```