#include <bits/stdc++.h>
using namespace std;
inline int in() {
  int res = 0;
  char c;
  while ((c = getchar()) < '0' || c > '9')
    ;
  while (c >= '0' && c <= '9') res = res * 10 + c - '0', c = getchar();
  return res;
}
const int N = 100010;
int a[N];
int dp[N];
int b[N];
int main() {
  int n = in(), T = in();
  int k = 0;
  for (int i = 0; i < n; i++) {
    a[i] = in();
    b[a[i]]++;
    k = max(k, b[a[i]]);
  }
  int t = min(n, T);
  for (int i = n; i < n * t; i++) {
    a[i] = a[i - n];
  }
  memset(dp, 0x3f3f3f3f, sizeof(dp));
  for (int i = 0; i < n * t; i++) {
    *upper_bound(dp, dp + n * t, a[i]) = a[i];
  }
  printf("%d\n", lower_bound(dp, dp + n * t, 0x3f3f3f3f) - dp + (T - t) * k);
  return 0;
}

### Prompt

Please provide a cpp coded solution 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;
inline int in() {
  int res = 0;
  char c;
  while ((c = getchar()) < '0' || c > '9')
    ;
  while (c >= '0' && c <= '9') res = res * 10 + c - '0', c = getchar();
  return res;
}
const int N = 100010;
int a[N];
int dp[N];
int b[N];
int main() {
  int n = in(), T = in();
  int k = 0;
  for (int i = 0; i < n; i++) {
    a[i] = in();
    b[a[i]]++;
    k = max(k, b[a[i]]);
  }
  int t = min(n, T);
  for (int i = n; i < n * t; i++) {
    a[i] = a[i - n];
  }
  memset(dp, 0x3f3f3f3f, sizeof(dp));
  for (int i = 0; i < n * t; i++) {
    *upper_bound(dp, dp + n * t, a[i]) = a[i];
  }
  printf("%d\n", lower_bound(dp, dp + n * t, 0x3f3f3f3f) - dp + (T - t) * k);
  return 0;
}

```