#include <bits/stdc++.h>
using namespace std;
int in() {
  int ret;
  cin >> ret;
  return ret;
}
const int N = 1e6 + 10, INF = 1e9 + 7;
int n, k, a[N], num[310], MAX, dp[310], ans;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    a[i] = in();
    num[a[i]]++;
    MAX = max(MAX, num[a[i]]);
  }
  for (int tmp = 1; tmp <= min(n, k); tmp++)
    for (int i = 1; i <= n; i++) a[i + (n * tmp)] = a[i];
  for (int i = 1; i <= (n * min(n, k)); i++) {
    dp[a[i]]++;
    for (int j = a[i] - 1; j >= 0; j--)
      dp[a[i]] = max(dp[a[i]], dp[j] + 1), ans = max(dp[a[i]], ans);
  }
  ans += MAX * (k - min(n, k));
  cout << ans << '\n';
  return 0;
}

### Prompt

Please formulate 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 in() {
  int ret;
  cin >> ret;
  return ret;
}
const int N = 1e6 + 10, INF = 1e9 + 7;
int n, k, a[N], num[310], MAX, dp[310], ans;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    a[i] = in();
    num[a[i]]++;
    MAX = max(MAX, num[a[i]]);
  }
  for (int tmp = 1; tmp <= min(n, k); tmp++)
    for (int i = 1; i <= n; i++) a[i + (n * tmp)] = a[i];
  for (int i = 1; i <= (n * min(n, k)); i++) {
    dp[a[i]]++;
    for (int j = a[i] - 1; j >= 0; j--)
      dp[a[i]] = max(dp[a[i]], dp[j] + 1), ans = max(dp[a[i]], ans);
  }
  ans += MAX * (k - min(n, k));
  cout << ans << '\n';
  return 0;
}

```