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 INF = 987654321;
int N, MAXB, T, B[1000010];
int max_by(int i, int *bit) {
  int s = 0;
  for (; i > 0; i -= i & -i) s = max(s, bit[i]);
  return s;
}
void increase(int i, int v, int *bit, int n) {
  for (; i <= n; i += i & -i) bit[i] = max(bit[i], v);
}
int bit[1000010], val[1000010];
void solve() {
  for (int i = 0; i < (int)(N); i++) scanf("%d", &B[i]);
  map<int, int> ma;
  for (int i = 0; i < (int)(N); i++) ma[B[i]] = 0;
  int K = 1;
  for (__typeof((ma).begin()) ite = (ma).begin(); ite != (ma).end(); ++ite)
    ite->second = K++;
  for (int i = 0; i < (int)(N); i++) B[i] = ma[B[i]];
  T = min(T, min(N, MAXB));
  memset(bit, 0, sizeof(bit));
  memset(val, 0, sizeof(val));
  for (int i = 0; i < (int)(T); i++)
    for (int i = 0; i < (int)(N); i++) {
      int t = 1 + max_by(B[i] - 1, bit);
      if (t > val[B[i]]) {
        increase(B[i], t, bit, N);
        val[B[i]] = t;
      }
    }
  printf("%d\n", max_by(N, bit));
}
int main() {
  int num_cases;
  while (4 == scanf("%d%d%d%d", &num_cases, &N, &MAXB, &T)) {
    for (int case_id = 0; case_id < (int)(num_cases); case_id++) solve();
  }
  return 0;
}
