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 lis(vector<int>& a) {
  if (a.empty()) return 0;
  int n = a.size();
  int INF = 1000;
  vector<int> b(n + 1, INF);
  b[0] = -INF;
  for (int i = 0; i < n; ++i) {
    int val = a[i];
    int pos = upper_bound(b.begin(), b.end(), val) - b.begin();
    b[pos] = min(b[pos], val);
  }
  int res = 0;
  for (int j = n; j >= 0; --j)
    if (b[j] != INF) {
      res = j;
      break;
    }
  return res;
}
int brute(int n, int t, vector<int> a) {
  for (int i = 0; i < t - 1; ++i)
    for (int j = 0; j < n; ++j) a.push_back(a[j]);
  int res = lis(a);
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  int n, t;
  cin >> n >> t;
  vector<int> a(n);
  for (int i = 0; i < n; ++i) cin >> a[i];
  int t0 = 200;
  int t1 = 201;
  int res = -1;
  if (t <= t0) {
    res = brute(n, t, a);
  } else {
    int res0 = brute(n, t0, a);
    int res1 = brute(n, t1, a);
    res = res1 + (t - t1) * (res1 - res0);
  }
  cout << res << endl;
  return 0;
}
