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 calc(vector<int> &arr) {
  vector<int> num(310, -2e+9);
  for (int iter = 0; iter < ((int)arr.size()); ++iter) {
    int ma = 0;
    for (int i = 0; i < arr[iter] + 1; ++i) {
      ma = max(ma, num[i]);
    }
    num[arr[iter]] = ma + 1;
  }
  int ans = 0;
  for (int iter = 0; iter < ((int)num.size()); ++iter)
    ans = max(ans, num[iter]);
  return ans;
}
int stupid(vector<int> &arr, int t, int n) {
  int last = 0;
  for (int iter = 0; iter < t; ++iter) {
    int cur = calc(arr);
    for (int i = 0; i < n; ++i) {
      arr.push_back(arr[i]);
    }
    last = cur;
  }
  return last;
}
int main() {
  int n, t;
  cin >> n >> t;
  vector<int> arr(n);
  for (int iter = 0; iter < n; ++iter) cin >> arr[iter];
  int last = 0;
  int dif = 0;
  int to = min(t, n);
  if (t <= 100) to = t;
  for (int iter = 0; iter < to; ++iter) {
    int cur = calc(arr);
    dif = cur - last;
    for (int i = 0; i < n; ++i) {
      arr.push_back(arr[i]);
    }
    last = cur;
  }
  int need = t - to;
  last = last + need * dif;
  cout << last << "\n";
  return 0;
}
