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, B, T, a[100005], p[100005], *f[100005], *q[100005];
int calc() {
  memset(p, 0, sizeof(p));
  for (int i = 1; i <= B; i++) f[1][i] = 1e9;
  for (int i = 1; i <= N; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= N; i++)
    if (!p[a[i]]) p[a[i]] = N + i, f[1][a[i]] = i;
  for (int i = N; i >= 0; i--) {
    for (int j = 1; j <= B; j++) q[i][j] = p[j];
    p[a[i]] = i;
  }
  for (int i = 2; i <= N; i++)
    for (int j = 1, m = 1e9, mm; j <= B; j++)
      mm = (m - 1) % N + 1, f[i][j] = q[mm][j] ? m + q[mm][j] - mm : 1e9,
      m = min(m, f[i - 1][j]);
  for (int i = N; i; i--)
    for (int j = 1; j <= B; j++)
      if (f[i][j] <= N * T) return i;
}
void doit() {
  scanf("%d%d%d%d", &K, &N, &B, &T), T = min(T, B);
  for (int i = 0; i <= N; i++) f[i] = new int[B + 5], q[i] = new int[B + 5];
  for (int i = 1; i <= K; i++) printf("%d\n", calc());
}
int main() {
  doit();
  return 0;
}
