#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e5 + 5;
const int MOD = 1e9 + 7;
const int INF = 2000000000;
int b[MAXN];
int main() {
  for (int i = 0; i < MAXN; i++) b[i] = INF;
  int n, T, a[105];
  scanf("%d%d", &n, &T);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  if (T < 200) {
    for (int times = 1; times <= T; times++) {
      for (int i = 0; i < n; i++) *upper_bound(b, b + 205 * n, a[i]) = a[i];
    }
    printf("%d", upper_bound(b, b + 205 * n, INF - 1) - b);
    return 0;
  }
  for (int times = 1; times <= 199; times++) {
    for (int i = 0; i < n; i++) *upper_bound(b, b + 205 * n, a[i]) = a[i];
  }
  int len1 = upper_bound(b, b + 205 * n, INF - 1) - b;
  for (int i = 0; i < n; i++) *upper_bound(b, b + 205 * n, a[i]) = a[i];
  int len2 = upper_bound(b, b + 205 * n, INF - 1) - b;
  printf("%d", len2 + (len2 - len1) * (T - 200));
  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;
const int MAXN = 1e5 + 5;
const int MOD = 1e9 + 7;
const int INF = 2000000000;
int b[MAXN];
int main() {
  for (int i = 0; i < MAXN; i++) b[i] = INF;
  int n, T, a[105];
  scanf("%d%d", &n, &T);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  if (T < 200) {
    for (int times = 1; times <= T; times++) {
      for (int i = 0; i < n; i++) *upper_bound(b, b + 205 * n, a[i]) = a[i];
    }
    printf("%d", upper_bound(b, b + 205 * n, INF - 1) - b);
    return 0;
  }
  for (int times = 1; times <= 199; times++) {
    for (int i = 0; i < n; i++) *upper_bound(b, b + 205 * n, a[i]) = a[i];
  }
  int len1 = upper_bound(b, b + 205 * n, INF - 1) - b;
  for (int i = 0; i < n; i++) *upper_bound(b, b + 205 * n, a[i]) = a[i];
  int len2 = upper_bound(b, b + 205 * n, INF - 1) - b;
  printf("%d", len2 + (len2 - len1) * (T - 200));
  return 0;
}

```