#include <bits/stdc++.h>
using namespace std;
int N;
int dp[302];
int ar[102];
int ans[302][302];
void update(int i, int j) {
  for (int a = i; a <= 300; a = a + (a & -a)) dp[a] = max(dp[a], j);
}
int q(int i) {
  int s = 0;
  for (int a = i; a >= 1; a = a - (a & -a)) s = max(s, dp[a]);
  return s;
}
int main() {
  int a, b, c, d, e, x, y, z, qu, k;
  scanf("%d %d", &N, &k);
  for (a = 0; a < N; a++) {
    scanf("%d", &ar[a]);
  }
  z = 0;
  for (e = 1; e <= min(N, k); e++) {
    for (a = 0; a < N; a++) {
      x = q(ar[a]);
      z = max(z, x + 1);
      update(ar[a], x + 1);
      ans[e][ar[a]] = max(ans[e][ar[a]], x + 1);
    }
  }
  if (k <= N) {
    cout << z << endl;
    return 0;
  }
  e = min(N, k);
  int n = k - e;
  for (a = 0; a < N; a++) {
    x = ans[e][ar[a]];
    y = x - ans[e - 1][ar[a]];
    z = max(z, x + n * y);
  }
  cout << z << endl;
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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;
int dp[302];
int ar[102];
int ans[302][302];
void update(int i, int j) {
  for (int a = i; a <= 300; a = a + (a & -a)) dp[a] = max(dp[a], j);
}
int q(int i) {
  int s = 0;
  for (int a = i; a >= 1; a = a - (a & -a)) s = max(s, dp[a]);
  return s;
}
int main() {
  int a, b, c, d, e, x, y, z, qu, k;
  scanf("%d %d", &N, &k);
  for (a = 0; a < N; a++) {
    scanf("%d", &ar[a]);
  }
  z = 0;
  for (e = 1; e <= min(N, k); e++) {
    for (a = 0; a < N; a++) {
      x = q(ar[a]);
      z = max(z, x + 1);
      update(ar[a], x + 1);
      ans[e][ar[a]] = max(ans[e][ar[a]], x + 1);
    }
  }
  if (k <= N) {
    cout << z << endl;
    return 0;
  }
  e = min(N, k);
  int n = k - e;
  for (a = 0; a < N; a++) {
    x = ans[e][ar[a]];
    y = x - ans[e - 1][ar[a]];
    z = max(z, x + n * y);
  }
  cout << z << endl;
  return 0;
}

```