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 f[20000007], maxc[100005], b[100005];
int tt, n, cs, t, maxb, len, ans;
void doit(int s, int t) {
  while ((s <= maxb) && (t > maxc[s])) {
    maxc[s] = t;
    ++s;
  }
}
int main() {
  scanf("%d %d %d %d", &tt, &n, &maxb, &t);
  t = min(t, maxb);
  for (int gb = 1; gb <= tt; gb++) {
    for (int i = 1; i <= n; i++) scanf("%d", &b[i]);
    len = t * n;
    memset(f, 0, sizeof(f));
    memset(maxc, 0, sizeof(maxc));
    f[1] = 1, doit(b[1], f[1]);
    ans = 1;
    for (int i = 2; i <= len; i++) {
      cs = b[(i - 1) % n + 1];
      f[i] = maxc[cs - 1] + 1;
      ans = max(ans, f[i]);
      doit(cs, f[i]);
    }
    printf("%d\n", ans);
  }
  return 0;
}
