Maxim loves sequences, especially those that strictly increase. He is wondering, what is the length of the longest increasing subsequence of the given sequence a?

Sequence a is given as follows: 

  * the length of the sequence equals n × t; 
  * <image> (1 ≤ i ≤ n × t), where operation <image> means taking the remainder after dividing number x by number y. 



Sequence s1, s2, ..., sr of length r is a subsequence of sequence a1, a2, ..., an, if there is such increasing sequence of indexes i1, i2, ..., ir (1 ≤ i1 < i2 < ... < ir ≤ n), that aij = sj. In other words, the subsequence can be obtained from the sequence by crossing out some elements.

Sequence s1, s2, ..., sr is increasing, if the following inequality holds: s1 < s2 < ... < sr.

Maxim have k variants of the sequence a. Help Maxim to determine for each sequence the length of the longest increasing subsequence.

Input

The first line contains four integers k, n, maxb and t (1 ≤ k ≤ 10; 1 ≤ n, maxb ≤ 105; 1 ≤ t ≤ 109; n × maxb ≤ 2·107). Each of the next k lines contain n integers b1, b2, ..., bn (1 ≤ bi ≤ maxb). 

Note that for each variant of the sequence a the values n, maxb and t coincide, the only arrays bs differ.

The numbers in the lines are separated by single spaces.

Output

Print k integers — the answers for the variants of the sequence a. Print the answers in the order the variants follow in the input.

Examples

Input

3 3 5 2
3 2 1
1 2 3
2 3 1


Output

2
3
3
#include <bits/stdc++.h>
using namespace std;
int len[100001];
int k, n, maxb, t;
int calc() {
  vector<int> inp;
  for (int i = 0; i < n; i++) {
    int m;
    cin >> m;
    inp.push_back(m);
  }
  vector<int> tmp = inp;
  sort(tmp.begin(), tmp.end());
  tmp.erase(unique(tmp.begin(), tmp.end()), tmp.end());
  if (tmp.size() <= t) {
    cout << tmp.size() << endl;
    return 0;
  }
  map<int, int> code;
  for (int i = 0; i < tmp.size(); i++) code[tmp[i]] = i;
  for (int i = 0; i < inp.size(); i++) {
    inp[i] = code[inp[i]];
  }
  for (int i = 0; i < 100001; i++) len[i] = 0;
  for (int i = 0; i < t; i++) {
    for (int j = 0; j < n; j++) {
      int newVal = 1;
      if (inp[j] > 0) newVal = len[inp[j] - 1] + 1;
      for (int k = inp[j]; k < tmp.size() && len[k] < newVal; k++)
        len[k] = newVal;
    }
  }
  int ret = 0;
  for (int i = 0; i < 100001; i++)
    if (len[i] > ret) ret = len[i];
  cout << ret << endl;
  return 0;
}
int main() {
  cin >> k >> n >> maxb >> t;
  for (int i = 0; i < k; i++) {
    calc();
  }
  return 0;
}
