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 N = 1e5 + 5;
int T, n, m, t, ans;
int a[N], c[N], dp[N];
int main() {
  scanf("%d%d%d%d", &T, &n, &m, &t);
  while (T--) {
    memset(c, 0, sizeof(c));
    memset(dp, 0, sizeof(dp));
    for (int i = 1; i <= n; i++) scanf("%d", a + i);
    if (t >= min(n, m)) {
      ans = 0;
      for (int i = 1; i <= n; i++)
        if (!c[a[i]]) {
          c[a[i]] = 1;
          ans++;
        }
      printf("%d\n", ans);
    } else {
      for (int i = 1; i <= t; i++)
        for (int j = 1; j <= n; j++) {
          int x = a[j];
          int val = dp[x - 1] + 1;
          for (int k = x; k <= m && val > dp[k]; k++) dp[k] = val;
        }
      printf("%d\n", dp[m]);
    }
  }
  return 0;
}
