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 m, n, mb, t;
int bb[2010000];
int has[100100];
int f[100100];
int main() {
  int i, oo, j, all;
  scanf("%d%d%d%d", &m, &n, &mb, &t);
  int ls = n;
  for (oo = 1; oo <= m; oo++) {
    int cnt = 0, ans = 0;
    memset(has, 0, sizeof(has));
    memset(f, 0, sizeof(f));
    for (i = 1; i <= n; i++) {
      scanf("%d", &bb[i]);
      if (has[bb[i]] == 0) cnt++;
      has[bb[i]]++;
    }
    if (t >= cnt) {
      cout << cnt << endl;
      continue;
    }
    all = 0;
    for (i = 1; i <= n * t; i++) {
      int ls = bb[(i - 1) % n + 1];
      int cur = f[ls - 1] + 1;
      ans = max(ans, cur);
      while (f[ls] < cur && ls <= mb) f[ls] = cur, ls++;
    }
    cout << ans << endl;
  }
  return 0;
}
