#include <bits/stdc++.h>
using namespace std;
const int N = 12345;
int a[N];
int f[N];
int dp[N];
int main() {
  int n, m, t;
  scanf("%d %d", &n, &t);
  m = n;
  for (int i = 0; i < n; i++) {
    scanf("%d", a + i);
    f[--a[i]]++;
  }
  int mx = 0;
  for (int i = 0; i < 300; i++) {
    mx = max(mx, f[i]);
  }
  for (int i = 0; i < min(m - 1, t - 1); i++) {
    for (int j = n; j < n + m; j++) {
      a[j] = a[j - m];
    }
    n += m;
  }
  int ans = 0;
  for (int i = 0; i < n; i++) {
    dp[i] = 1;
    for (int j = i - 1; j >= 0; j--) {
      if (a[i] >= a[j]) {
        dp[i] = max(dp[i], dp[j] + 1);
      }
    }
    ans = max(ans, dp[i]);
  }
  t -= min(m, t);
  ans += t * mx;
  printf("%d\n", ans);
  return 0;
}

### 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;
const int N = 12345;
int a[N];
int f[N];
int dp[N];
int main() {
  int n, m, t;
  scanf("%d %d", &n, &t);
  m = n;
  for (int i = 0; i < n; i++) {
    scanf("%d", a + i);
    f[--a[i]]++;
  }
  int mx = 0;
  for (int i = 0; i < 300; i++) {
    mx = max(mx, f[i]);
  }
  for (int i = 0; i < min(m - 1, t - 1); i++) {
    for (int j = n; j < n + m; j++) {
      a[j] = a[j - m];
    }
    n += m;
  }
  int ans = 0;
  for (int i = 0; i < n; i++) {
    dp[i] = 1;
    for (int j = i - 1; j >= 0; j--) {
      if (a[i] >= a[j]) {
        dp[i] = max(dp[i], dp[j] + 1);
      }
    }
    ans = max(ans, dp[i]);
  }
  t -= min(m, t);
  ans += t * mx;
  printf("%d\n", ans);
  return 0;
}

```