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 n, t, arr[101], memo[100000];
int freq[305];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> t;
  int mx = 0;
  for (int i = 0; i < (long long)(n); i++)
    cin >> arr[i], mx = max(mx, ++freq[arr[i]]);
  vector<int> vec;
  int nn = n * n;
  for (int i = 0; i < (long long)(nn); i++) {
    int j = i % n;
    auto it = upper_bound(((vec).begin()), ((vec).end()), arr[j]);
    if (it == vec.end())
      vec.push_back(arr[j]);
    else
      *it = arr[j];
    memo[i] = ((int)((vec).size()));
  }
  int hoba = n * t;
  if (hoba > nn)
    cout << memo[nn - 1] + mx * (t - n);
  else
    cout << memo[hoba - 1] << '\n';
  return 0;
}
