#include <bits/stdc++.h>
using namespace std;
const int N = 105;
int a[N], sq[N * N], seq[N * N], prf[N * N], tp = 0, mct = 0, cnt[N], n, l;
int main() {
  scanf("%d%d", &n, &l);
  for (int i = 1; i <= n; i++) scanf("%d", a + i);
  for (int i = 0; i < min(n, l); i++) {
    memcpy(sq + 1 + i * n, a + 1, n * sizeof(int));
  }
  sort(a + 1, a + n + 1);
  for (int p, i = 1; i <= n; i++) {
    p = lower_bound(a + 1, a + n + 1, a[i]) - a;
    mct = max(mct, ++cnt[p]);
  }
  for (int p, i = 1; i <= n * min(n, l); i++) {
    if (sq[i] >= seq[tp]) {
      prf[i] = ++tp;
      seq[tp] = sq[i];
    } else {
      p = upper_bound(seq + 1, seq + tp + 1, sq[i]) - seq;
      while (seq[p] == sq[i]) p++;
      prf[i] = p;
      seq[p] = sq[i];
    }
  }
  if (n >= l) {
    cout << tp << endl;
  } else {
    cout << tp + (l - n) * mct << endl;
  }
}

### Prompt

Develop a solution in CPP 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 = 105;
int a[N], sq[N * N], seq[N * N], prf[N * N], tp = 0, mct = 0, cnt[N], n, l;
int main() {
  scanf("%d%d", &n, &l);
  for (int i = 1; i <= n; i++) scanf("%d", a + i);
  for (int i = 0; i < min(n, l); i++) {
    memcpy(sq + 1 + i * n, a + 1, n * sizeof(int));
  }
  sort(a + 1, a + n + 1);
  for (int p, i = 1; i <= n; i++) {
    p = lower_bound(a + 1, a + n + 1, a[i]) - a;
    mct = max(mct, ++cnt[p]);
  }
  for (int p, i = 1; i <= n * min(n, l); i++) {
    if (sq[i] >= seq[tp]) {
      prf[i] = ++tp;
      seq[tp] = sq[i];
    } else {
      p = upper_bound(seq + 1, seq + tp + 1, sq[i]) - seq;
      while (seq[p] == sq[i]) p++;
      prf[i] = p;
      seq[p] = sq[i];
    }
  }
  if (n >= l) {
    cout << tp << endl;
  } else {
    cout << tp + (l - n) * mct << endl;
  }
}

```