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. 
#include <bits/stdc++.h>
using namespace std;
vector<int> s, ss;
map<int, int> mmap;
map<int, int>::iterator it;
int dp[111111];
int main() {
  int n, t;
  cin >> n >> t;
  for (int i = 0; i < n; i++) {
    int mid;
    cin >> mid;
    ss.push_back(mid);
  }
  for (int i = 0; i < n; i++) {
    mmap[ss[i]]++;
  }
  int ansnum = 0;
  for (int i = 1; i <= min(n, t); i++) {
    for (int k = 0; k < n; k++) s.push_back(ss[k]);
  }
  int len = s.size();
  int ans = 0;
  for (int i = 0; i < len; i++) {
    dp[i] = 1;
    for (int j = i - 1; j >= max(i - n, 0); j--) {
      if (s[i] >= s[j]) {
        dp[i] = max(dp[i], dp[j] + 1);
      }
    }
    ans = max(ans, dp[i]);
  }
  int maxn = 0;
  for (it = mmap.begin(); it != mmap.end(); it++) {
    maxn = max(maxn, it->second);
  }
  cout << max(ans, ans + (t - n) * maxn) << endl;
}
