#include <bits/stdc++.h>
using namespace std;
const int inf = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const long long mmax = 1000;
int a[500];
int b[500000];
int dp[500000];
int cnt[500];
int main() {
  memset(dp, inf, sizeof dp);
  int n, t;
  cin >> n >> t;
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    cnt[a[i]]++;
  }
  int x;
  if (t >= 1000)
    x = 1000;
  else
    x = t;
  for (int i = 0; i < n * x; i++) {
    b[i + 1] = a[i % n];
  }
  int nmin = 0;
  for (int i = 0; i <= 300; i++) {
    nmin = max(nmin, cnt[i]);
  }
  int ans = 1;
  for (int i = 1; i <= n * x; i++) {
    int l = 1, r = ans;
    while (l <= r) {
      int mid = (l + r) >> 1;
      if (dp[mid] <= b[i]) {
        l = mid + 1;
      } else
        r = mid - 1;
    }
    dp[l] = b[i];
    ans = max(ans, l);
  }
  if (t >= 1000) {
    cout << (long long)ans + (long long)(nmin) * (t - 1000) << endl;
  } else
    cout << ans << endl;
  return 0;
}

### Prompt

Your task is to create a Cpp solution 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;
const int inf = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const long long mmax = 1000;
int a[500];
int b[500000];
int dp[500000];
int cnt[500];
int main() {
  memset(dp, inf, sizeof dp);
  int n, t;
  cin >> n >> t;
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    cnt[a[i]]++;
  }
  int x;
  if (t >= 1000)
    x = 1000;
  else
    x = t;
  for (int i = 0; i < n * x; i++) {
    b[i + 1] = a[i % n];
  }
  int nmin = 0;
  for (int i = 0; i <= 300; i++) {
    nmin = max(nmin, cnt[i]);
  }
  int ans = 1;
  for (int i = 1; i <= n * x; i++) {
    int l = 1, r = ans;
    while (l <= r) {
      int mid = (l + r) >> 1;
      if (dp[mid] <= b[i]) {
        l = mid + 1;
      } else
        r = mid - 1;
    }
    dp[l] = b[i];
    ans = max(ans, l);
  }
  if (t >= 1000) {
    cout << (long long)ans + (long long)(nmin) * (t - 1000) << endl;
  } else
    cout << ans << endl;
  return 0;
}

```