#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e5 * 5;
const int base = 1e9 + 7;
const int N = 50000;
const double pi = acos(-1.0);
const double ep = 1e-6;
int n, t;
int ans;
int a[N];
int d[N], dp[N];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> t;
  for (int i = 1; i <= n; i++) cin >> a[i];
  if (t <= 2 * n) {
    for (int i = n + 1; i <= t * n; i++) a[i] = a[i - n];
    for (int i = 1; i <= t * n; i++) {
      for (int j = a[i]; j >= 1; j--) dp[a[i]] = max(dp[a[i]], dp[j] + 1);
    }
    for (int i = 1; i <= 300; i++) ans = max(ans, dp[i]);
    cout << ans << "\n";
  } else {
    int p = n * n;
    for (int i = n + 1; i <= p; i++) a[i] = a[i - n];
    for (int i = 1; i <= n; i++) ++d[a[i]];
    for (int i = 1; i <= p; i++)
      for (int j = a[i]; j >= 1; j--) dp[a[i]] = max(dp[a[i]], dp[j] + 1);
    for (int i = 1; i <= 300; i++) dp[i] = dp[i] + d[i] * (t - 2 * n);
    for (int i = 1; i <= p; i++)
      for (int j = a[i]; j >= 1; j--) dp[a[i]] = max(dp[a[i]], dp[j] + 1);
    for (int i = 1; i <= 300; i++) ans = max(ans, dp[i]);
    cout << ans << "\n";
  }
  return 0;
}

### Prompt

Please formulate 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 MAXN = 1e5 * 5;
const int base = 1e9 + 7;
const int N = 50000;
const double pi = acos(-1.0);
const double ep = 1e-6;
int n, t;
int ans;
int a[N];
int d[N], dp[N];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> t;
  for (int i = 1; i <= n; i++) cin >> a[i];
  if (t <= 2 * n) {
    for (int i = n + 1; i <= t * n; i++) a[i] = a[i - n];
    for (int i = 1; i <= t * n; i++) {
      for (int j = a[i]; j >= 1; j--) dp[a[i]] = max(dp[a[i]], dp[j] + 1);
    }
    for (int i = 1; i <= 300; i++) ans = max(ans, dp[i]);
    cout << ans << "\n";
  } else {
    int p = n * n;
    for (int i = n + 1; i <= p; i++) a[i] = a[i - n];
    for (int i = 1; i <= n; i++) ++d[a[i]];
    for (int i = 1; i <= p; i++)
      for (int j = a[i]; j >= 1; j--) dp[a[i]] = max(dp[a[i]], dp[j] + 1);
    for (int i = 1; i <= 300; i++) dp[i] = dp[i] + d[i] * (t - 2 * n);
    for (int i = 1; i <= p; i++)
      for (int j = a[i]; j >= 1; j--) dp[a[i]] = max(dp[a[i]], dp[j] + 1);
    for (int i = 1; i <= 300; i++) ans = max(ans, dp[i]);
    cout << ans << "\n";
  }
  return 0;
}

```