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 i, j, k, l, s, n, m, test, maxb, t, F[100005], a[100005], f[100005];
inline void change(int o, int p) {
  for (; o <= 100005 - 5; o += o & -o) f[o] = max(f[o], p);
}
inline int ask(int o) {
  int s = 0;
  for (; o; o -= o & -o) s = max(s, f[o]);
  return s;
}
int main() {
  scanf("%d%d%d%d", &test, &n, &maxb, &t);
  t = min(t, maxb);
  while (test--) {
    for (i = 1; i <= n; i++) scanf("%d", &a[i]), F[i] = 0;
    memset(f, 0, sizeof(f));
    for (i = 1; i <= t; i++)
      for (j = 1; j <= n; j++) {
        int gt = ask(a[j]);
        if (gt >= F[j]) change(a[j] + 1, gt + 1), F[j] = gt + 1;
      }
    printf("%d\n", ask(100005 - 5));
  }
}
