#include <bits/stdc++.h>
using namespace std;
const int N = 3e5 + 7;
const int M = 1e6 + 7;
const int lim = 2e6;
const int P = 998244353;
const int inf = 0x3f3f3f3f;
int n, k, mx;
int a[N], b[N], dp[N];
map<int, int> mp;
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    mp[a[i]]++;
    mx = max(mx, mp[a[i]]);
  }
  int m = min(k, n), len = n, t = 0;
  dp[0] = -1;
  for (int i = 2; i <= m; i++)
    for (int j = 1; j <= n; j++) a[++len] = a[j];
  for (int i = 1; i <= len; i++) {
    dp[i] = 1;
    for (int j = 1; j < i; j++) {
      if (a[i] >= a[j]) dp[i] = max(dp[i], dp[j] + 1);
    }
    t = max(t, dp[i]);
  }
  printf("%d\n", t + (k - m) * mx);
  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;
const int N = 3e5 + 7;
const int M = 1e6 + 7;
const int lim = 2e6;
const int P = 998244353;
const int inf = 0x3f3f3f3f;
int n, k, mx;
int a[N], b[N], dp[N];
map<int, int> mp;
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    mp[a[i]]++;
    mx = max(mx, mp[a[i]]);
  }
  int m = min(k, n), len = n, t = 0;
  dp[0] = -1;
  for (int i = 2; i <= m; i++)
    for (int j = 1; j <= n; j++) a[++len] = a[j];
  for (int i = 1; i <= len; i++) {
    dp[i] = 1;
    for (int j = 1; j < i; j++) {
      if (a[i] >= a[j]) dp[i] = max(dp[i], dp[j] + 1);
    }
    t = max(t, dp[i]);
  }
  printf("%d\n", t + (k - m) * mx);
  return 0;
}

```