#include <bits/stdc++.h>
using namespace std;
int n, t;
int a[10005];
int b[505];
int dp[10005];
int main() {
  scanf("%d%d", &n, &t);
  int len = n * min(n, t);
  for (register int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    b[a[i]]++;
  }
  int maxn = 1;
  for (int i = n + 1; i <= len; i++) {
    a[i] = a[i - n];
  }
  for (int i = 1; i <= len; i++) {
    dp[i] = 1;
    for (int j = i - 1; j >= max(i - n, 1); j--) {
      if (a[i] >= a[j]) dp[i] = max(dp[i], dp[j] + 1);
    }
    maxn = max(maxn, dp[i]);
  }
  if (n >= t)
    printf("%d", maxn);
  else {
    int maxnn = 0;
    for (int i = 0; i < 505; i++) {
      maxnn = max(maxnn, b[i]);
    }
    printf("%d", maxn + maxnn * (t - n));
  }
  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;
int n, t;
int a[10005];
int b[505];
int dp[10005];
int main() {
  scanf("%d%d", &n, &t);
  int len = n * min(n, t);
  for (register int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    b[a[i]]++;
  }
  int maxn = 1;
  for (int i = n + 1; i <= len; i++) {
    a[i] = a[i - n];
  }
  for (int i = 1; i <= len; i++) {
    dp[i] = 1;
    for (int j = i - 1; j >= max(i - n, 1); j--) {
      if (a[i] >= a[j]) dp[i] = max(dp[i], dp[j] + 1);
    }
    maxn = max(maxn, dp[i]);
  }
  if (n >= t)
    printf("%d", maxn);
  else {
    int maxnn = 0;
    for (int i = 0; i < 505; i++) {
      maxnn = max(maxnn, b[i]);
    }
    printf("%d", maxn + maxnn * (t - n));
  }
  return 0;
}

```