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 k, n, maxb, t, sum, b[100005], a[100005], f[100005], ans;
bool c[100005];
void add(int x, int y) {
  for (; x <= maxb; x += x & (-x)) a[x] = max(a[x], y);
}
int ask(int x) {
  int p = 0;
  for (; x; x -= x & (-x)) p = max(p, a[x]);
  return p;
}
void work() {
  memset(c, 0, sizeof(c));
  memset(f, 0, sizeof(f));
  memset(a, 0, sizeof(a));
  sum = ans = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &b[i]);
    if (!c[b[i]]) c[b[i]] = 1, sum++;
  }
  if (t >= sum) {
    printf("%d\n", sum);
    return;
  }
  for (int i = 1; i <= t; i++)
    for (int j = 1; j <= n; j++) {
      int o = ask(b[j] - 1) + 1;
      if (o > f[j]) {
        f[j] = o;
        add(b[j], o);
        ans = max(ans, o);
      }
      if (ans >= sum) break;
    }
  printf("%d\n", ans);
}
int main() {
  scanf("%d%d%d%d", &k, &n, &maxb, &t);
  for (int i = 1; i <= k; i++) work();
  return 0;
}
