#include <bits/stdc++.h>
using namespace std;
long long n, T, num[305], b[10005], a[10005], ans, cnt, len;
int main() {
  scanf("%d%d", &n, &T);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
    num[a[i]]++;
    ans = max(ans, num[a[i]]);
  }
  if (T >= n)
    ans *= (T - n);
  else
    ans = 0;
  for (int i = 1; i < n; ++i)
    for (int j = 1; j <= n; ++j) a[i * n + j] = a[j];
  if (T >= n)
    len = n * n;
  else
    len = n * T;
  for (int i = 1; i <= len; ++i) {
    int l = 1, r = cnt, mid;
    while (l <= r) {
      mid = (l + r) >> 1;
      if (b[mid] <= a[i])
        l = mid + 1;
      else
        r = mid - 1;
    }
    if (l > cnt) ++cnt;
    b[l] = a[i];
  }
  printf("%d\n", ans + cnt);
  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;
long long n, T, num[305], b[10005], a[10005], ans, cnt, len;
int main() {
  scanf("%d%d", &n, &T);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
    num[a[i]]++;
    ans = max(ans, num[a[i]]);
  }
  if (T >= n)
    ans *= (T - n);
  else
    ans = 0;
  for (int i = 1; i < n; ++i)
    for (int j = 1; j <= n; ++j) a[i * n + j] = a[j];
  if (T >= n)
    len = n * n;
  else
    len = n * T;
  for (int i = 1; i <= len; ++i) {
    int l = 1, r = cnt, mid;
    while (l <= r) {
      mid = (l + r) >> 1;
      if (b[mid] <= a[i])
        l = mid + 1;
      else
        r = mid - 1;
    }
    if (l > cnt) ++cnt;
    b[l] = a[i];
  }
  printf("%d\n", ans + cnt);
  return 0;
}

```