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;
const int Maxn = 100005;
int k, n, t, tot, a[Maxn], b[Maxn], f[Maxn];
inline int get() {
  char ch;
  while (!isdigit(ch = getchar()))
    ;
  int v = ch - 48;
  while (isdigit(ch = getchar())) v = v * 10 + ch - 48;
  return v;
}
int main() {
  k = get();
  n = get();
  get();
  t = get();
  while (k--) {
    for (int i = 1; i <= n; i++) a[i] = b[i] = get();
    sort(b + 1, b + n + 1);
    tot = unique(b + 1, b + n + 1) - b - 1;
    for (int i = 1; i <= n; i++)
      a[i] = lower_bound(b + 1, b + tot + 1, a[i]) - b;
    if (t >= tot) {
      printf("%d\n", tot);
      continue;
    }
    memset(f + 1, 0, (++tot) << 2);
    for (int i = 1, j = 1; i <= n * t; i++, (++j) > n ? j = 1 : 0)
      for (int u = a[j] + 1, v = f[a[j]] + 1; u <= tot && f[u] < v; u++)
        f[u] = v;
    printf("%d\n", f[tot]);
  }
  return 0;
}
