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 = 500010;
int k, n, maxb, t, b[N], tot = 1, dp[N << 2], rec[N << 2];
void update(int i, int x) {
  while (i <= tot) {
    dp[i] = max(dp[i], x);
    i += i & -i;
  }
}
int sum(int i) {
  int ret = 0;
  while (i) {
    ret = max(ret, dp[i]);
    i -= i & -i;
  }
  return ret;
}
int main() {
  scanf("%d%d%d%d", &k, &n, &maxb, &t);
  t = min(t, maxb);
  while (tot <= maxb) tot <<= 1;
  while (k--) {
    memset(dp, 0, sizeof(dp));
    memset(rec, 0, sizeof(rec));
    memset(b, 0, sizeof(b));
    for (int i = 0; i < n; i++) scanf("%d", &b[i]);
    for (int i = 1; i <= t; i++)
      for (int j = 0; j < n; j++) {
        int tmp = sum(b[j] - 1);
        if (tmp >= rec[j]) {
          update(b[j], tmp + 1);
          rec[j] = tmp + 1;
        }
      }
    printf("%d\n", sum(tot));
  }
  return 0;
}
