#include <bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:1024000000,1024000000")
int c[520], a[520 * 520], d[520 * 520];
int main() {
  int n, t;
  scanf("%d%d", &n, &t);
  int mx = 0;
  for (int i = 0; i < n; ++i) {
    scanf("%d", &a[i]);
    c[a[i]]++;
  }
  int cnt = 0;
  for (int i = 0; i < 520; ++i) {
    if (c[a[i]] > cnt) cnt = c[a[i]];
  }
  int tmp = min(t, n);
  for (int i = 1; i < tmp; ++i)
    for (int j = 0; j < n; ++j) a[i * n + j] = a[j];
  int L = n * tmp, len = 1;
  d[1] = a[0];
  int j;
  for (int i = 1; i < L; ++i) {
    if (d[len] <= a[i])
      j = ++len;
    else {
      j = upper_bound(d + 1, d + 1 + len, a[i]) - d;
    }
    d[j] = a[i];
  }
  cout << len + max(0, t - tmp) * cnt << endl;
  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;
#pragma comment(linker, "/STACK:1024000000,1024000000")
int c[520], a[520 * 520], d[520 * 520];
int main() {
  int n, t;
  scanf("%d%d", &n, &t);
  int mx = 0;
  for (int i = 0; i < n; ++i) {
    scanf("%d", &a[i]);
    c[a[i]]++;
  }
  int cnt = 0;
  for (int i = 0; i < 520; ++i) {
    if (c[a[i]] > cnt) cnt = c[a[i]];
  }
  int tmp = min(t, n);
  for (int i = 1; i < tmp; ++i)
    for (int j = 0; j < n; ++j) a[i * n + j] = a[j];
  int L = n * tmp, len = 1;
  d[1] = a[0];
  int j;
  for (int i = 1; i < L; ++i) {
    if (d[len] <= a[i])
      j = ++len;
    else {
      j = upper_bound(d + 1, d + 1 + len, a[i]) - d;
    }
    d[j] = a[i];
  }
  cout << len + max(0, t - tmp) * cnt << endl;
  return 0;
}

```