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 MAXN = (int)1e5 + 10;
int n, t, a[MAXN], am, p[MAXN];
void init() {
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  memset(p, 0, sizeof(p));
}
void solve() {
  int Tohka = 0;
  for (int i = 1; i <= n * t; ++i) {
    int w = a[(i - 1) % n + 1], f = p[w] + 1;
    for (int j = w + 1; p[j] < f && j <= am + 1; p[j] = f, ++j)
      ;
  }
  Tohka = p[am + 1];
  cout << Tohka << endl;
}
int main() {
  int test;
  scanf("%d%d%d%d", &test, &n, &am, &t);
  t = min(t, am);
  while (test) init(), solve(), --test;
  fclose(stdin);
  fclose(stdout);
  return 0;
}
