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 a[100011], b[100011], f[100011];
map<int, int> Hash;
int tj, dt, n, m, T, t, tot, i, j, tmp, ans;
int main() {
  scanf("%d%d%d%d", &dt, &n, &m, &T);
  for (tj = 1; tj <= dt; tj++) {
    Hash.clear();
    t = 0;
    for (i = 1; i <= n; i++) {
      scanf("%d", &a[i]);
      if (!Hash[a[i]]) b[++t] = a[i], Hash[a[i]] = 1;
    }
    sort(b + 1, b + 1 + t);
    if (T >= t)
      printf("%d\n", t);
    else {
      for (i = 1; i <= t; i++) Hash[b[i]] = i;
      for (i = 1; i <= n; i++) a[i] = Hash[a[i]];
      ans = 0;
      memset(f, 0, sizeof(f));
      memset(b, 0, sizeof(b));
      for (i = 1; i <= T; i++) {
        for (j = 1; j <= n; j++) {
          while (f[a[j]] <= ans && b[f[a[j]]] < a[j]) f[a[j]]++;
          b[f[a[j]]] = a[j];
          ans = max(ans, f[a[j]]);
          if (ans == t) break;
        }
        if (ans == t) break;
      }
      printf("%d\n", ans);
    }
  }
}
