#include <bits/stdc++.h>
using namespace std;
const int MAXN = 5010;
int n, t, a[MAXN], lim, f[MAXN], ans = 0, mf = 0, b[350] = {0};
int main() {
  scanf("%d%d", &n, &t);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    b[a[i]]++;
    mf = max(mf, b[a[i]]);
  }
  lim = n * min(n / 2, t);
  if (n == 1) lim = n;
  for (int i = n + 1; i <= lim; i++) a[i] = a[i - n];
  for (int i = 1; i <= lim; i++) {
    f[i] = 1;
    for (int j = 1; j <= i - 1; j++)
      if (a[j] <= a[i]) f[i] = max(f[i], f[j] + 1);
    ans = max(ans, f[i]);
  }
  if (n == 1) n++;
  if (t <= (n / 2))
    printf("%d\n", ans);
  else
    printf("%d\n", ans + (t - (n / 2)) * mf);
  return 0;
}

### Prompt

Create a solution in cpp for 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 MAXN = 5010;
int n, t, a[MAXN], lim, f[MAXN], ans = 0, mf = 0, b[350] = {0};
int main() {
  scanf("%d%d", &n, &t);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    b[a[i]]++;
    mf = max(mf, b[a[i]]);
  }
  lim = n * min(n / 2, t);
  if (n == 1) lim = n;
  for (int i = n + 1; i <= lim; i++) a[i] = a[i - n];
  for (int i = 1; i <= lim; i++) {
    f[i] = 1;
    for (int j = 1; j <= i - 1; j++)
      if (a[j] <= a[i]) f[i] = max(f[i], f[j] + 1);
    ans = max(ans, f[i]);
  }
  if (n == 1) n++;
  if (t <= (n / 2))
    printf("%d\n", ans);
  else
    printf("%d\n", ans + (t - (n / 2)) * mf);
  return 0;
}

```