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 T, n, maxb, t, a[100010], tr[100010], f[100010];
void put(int x, int y) {
  while (x <= maxb) {
    tr[x] = max(tr[x], y);
    x += x & (-x);
  }
}
int get(int x) {
  int sum = 0;
  while (x) {
    sum = max(sum, tr[x]);
    x -= x & (-x);
  }
  return sum;
}
int main() {
  scanf("%d%d%d%d", &T, &n, &maxb, &t);
  t = min(maxb, t);
  while (T--) {
    for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
    if (t >= maxb) {
      sort(a + 1, a + n + 1);
      printf("%d\n", unique(a + 1, a + n + 1) - (a + 1));
      continue;
    }
    memset(tr, 0, sizeof(tr));
    memset(f, 0, sizeof(f));
    for (int i = 1; i <= n * t; ++i) {
      int c = a[(i - 1) % n + 1];
      int tmp = get(c - 1) + 1;
      if (tmp > f[c]) f[c] = tmp, put(c, tmp);
    }
    printf("%d\n", get(maxb));
  }
  return 0;
}
