#include <bits/stdc++.h>
using namespace std;
int n, t;
int a[10002];
int frequency[400];
int dp[10005];
int maxCount;
int main() {
  scanf("%d %d", &n, &t);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    frequency[a[i]]++;
    maxCount = max(maxCount, frequency[a[i]]);
  }
  for (int i = n + 1; i <= n * min(t, n); i++) {
    int pos = i % n;
    if (pos == 0) pos = n;
    a[i] = a[pos];
  }
  int res = 1;
  for (int i = 1; i <= n * min(t, n); i++) {
    dp[i] = 1;
    for (int j = 0; j < i; j++) {
      if (a[j] <= a[i]) {
        dp[i] = max(dp[i], dp[j] + 1);
      }
      res = max(res, dp[i]);
    }
  }
  if (t > n) {
    res += maxCount * (t - n);
  }
  printf("%d\n", res);
  return 0;
}

### Prompt

Please create a solution in cpp to 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;
int n, t;
int a[10002];
int frequency[400];
int dp[10005];
int maxCount;
int main() {
  scanf("%d %d", &n, &t);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    frequency[a[i]]++;
    maxCount = max(maxCount, frequency[a[i]]);
  }
  for (int i = n + 1; i <= n * min(t, n); i++) {
    int pos = i % n;
    if (pos == 0) pos = n;
    a[i] = a[pos];
  }
  int res = 1;
  for (int i = 1; i <= n * min(t, n); i++) {
    dp[i] = 1;
    for (int j = 0; j < i; j++) {
      if (a[j] <= a[i]) {
        dp[i] = max(dp[i], dp[j] + 1);
      }
      res = max(res, dp[i]);
    }
  }
  if (t > n) {
    res += maxCount * (t - n);
  }
  printf("%d\n", res);
  return 0;
}

```