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 + 5;
int k, n, maxb, t;
int b[maxn], c[maxn], a[maxn], f[maxn];
void solved(int nT) {
  for (int i = 0; i < n; ++i) {
    scanf("%d", &b[i]);
    a[i] = b[i];
  }
  sort(a, a + n);
  int tn = unique(a, a + n) - a;
  if (t >= tn) {
    cout << tn << endl;
    return;
  }
  for (int i = 0; i < n; ++i) {
    b[i] = lower_bound(a, a + tn, b[i]) - a;
  }
  memset(f, 0, tn << 2);
  for (int ct = 0; ct < t; ++ct) {
    for (int i = 0; i < n; ++i) {
      int tmp = (b[i] == 0) ? 0 : f[b[i] - 1];
      int j = b[i];
      while (j < tn && f[j] == tmp) {
        ++f[j++];
      }
    }
  }
  cout << f[tn - 1] << endl;
}
int main() {
  scanf("%d %d %d %d", &k, &n, &maxb, &t);
  for (int nT = 1; nT <= k; ++nT) {
    solved(nT);
  }
  return 0;
}
