#include <bits/stdc++.h>
long long n, t, ans, maxn, a[105], cnt[305], dp[105];
long long _max(long long a, long long b) { return a > b ? a : b; }
int main() {
  scanf("%lld%lld", &n, &t);
  for (int i = 1; i <= n; ++i) {
    scanf("%lld", a + i);
    ++cnt[a[i]];
    maxn = _max(maxn, cnt[a[i]]);
  }
  for (int i = 1; i <= n && i <= t; ++i) {
    for (int j = 1; j <= n; ++j) {
      ++dp[j];
      for (int k = 1; k <= n; ++k) {
        if (j ^ k && a[j] >= a[k]) {
          dp[j] = _max(dp[j], dp[k] + 1);
        }
      }
    }
  }
  for (int i = 1; i <= n; ++i) {
    ans = _max(ans, dp[i]);
  }
  printf("%lld\n", ans + (t > n) * (t - n) * maxn);
  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>
long long n, t, ans, maxn, a[105], cnt[305], dp[105];
long long _max(long long a, long long b) { return a > b ? a : b; }
int main() {
  scanf("%lld%lld", &n, &t);
  for (int i = 1; i <= n; ++i) {
    scanf("%lld", a + i);
    ++cnt[a[i]];
    maxn = _max(maxn, cnt[a[i]]);
  }
  for (int i = 1; i <= n && i <= t; ++i) {
    for (int j = 1; j <= n; ++j) {
      ++dp[j];
      for (int k = 1; k <= n; ++k) {
        if (j ^ k && a[j] >= a[k]) {
          dp[j] = _max(dp[j], dp[k] + 1);
        }
      }
    }
  }
  for (int i = 1; i <= n; ++i) {
    ans = _max(ans, dp[i]);
  }
  printf("%lld\n", ans + (t > n) * (t - n) * maxn);
  return 0;
}

```