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;
int solve(vector<int> &v) {
  int n = v.size();
  vector<int> d(v.size());
  for (int i = 0; i < n; i++) {
    d[i] = 1;
    for (int j = 0; j < i; j++) {
      if (v[j] <= v[i]) {
        d[i] = max(d[i], 1 + d[j]);
      }
    }
  }
  int res = d[0];
  for (int i = 0; i < n; i++) res = max(res, d[i]);
  return res;
}
int main() {
  long long n, t;
  cin >> n >> t;
  vector<int> v(n);
  for (int i = 0; i < n; i++) {
    cin >> v[i];
  }
  if (t < 150) {
    vector<int> v1(n * t);
    for (int i = 0; i < v1.size(); i++) v1[i] = v[i % n];
    cout << solve(v1) << endl;
    return 0;
  }
  vector<int> v1(n * (100 + t % 2));
  for (int i = 0; i < v1.size(); i++) v1[i] = v[i % n];
  vector<int> v2(n * (102 + t % 2));
  for (int i = 0; i < v2.size(); i++) v2[i] = v[i % n];
  long long s1 = solve(v1);
  long long s2 = solve(v2);
  long long ds = s2 - s1;
  long long cnt = (n * t - v2.size()) / 2 / n;
  long long res = s2 + ds * cnt;
  cout << res << endl;
  return 0;
}
