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;
const int max_n = 1e3;
int n, T;
vector<int> vec(max_n), sec;
int d[max_n], c[max_n];
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0);
  cin >> n >> T;
  for (int i = 0; i < n; i++) cin >> vec[i];
  for (int i = 0; i < min(T, (int)8e4); i++) {
    for (int j = 0; j < n; j++) {
      int pos = upper_bound(sec.begin(), sec.end(), vec[j]) - sec.begin();
      if (pos == sec.size())
        sec.push_back(vec[j]);
      else
        sec[pos] = vec[j];
      c[j] = (pos + 1) - d[j];
      d[j] = pos + 1;
    }
  }
  int maxi = 0, ind = -1;
  for (int i = 0; i < n; i++) {
    if (maxi < d[i]) {
      maxi = d[i];
      ind = i;
    }
  }
  for (int i = min((int)8e4, T); i < T; i++) maxi += c[ind];
  cout << maxi << endl;
  return 0;
}
