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 = 100020;
const int MOd = 1e9 + 7;
int k, n, maxb, t, a;
int ar[maxn];
int segment[maxn * 3], fw[maxn];
int **f;
bool cmp(const pair<int, int> &a, const pair<int, int> &b) {
  if (a.first == b.first) return a.second < b.second;
  return a.first > b.first;
}
void solve() {
  for (int i = 1; i <= n; i++) scanf("%d", &ar[i]);
  memset(fw, 0, sizeof(fw));
  for (int i = t; i >= 1; i--) {
    for (int j = 1; j <= n; j++) {
      int t = 0;
      for (int k = ar[j] - 1; k > 0; k -= k & (-k)) t = max(t, (fw[k]));
      if (t + 1 > fw[ar[j]]) {
        for (int k = ar[j]; k <= maxb; k += k & (-k))
          fw[k] = max(fw[k], (t + 1));
      }
    }
  }
  int ret = 0;
  for (int i = 1; i <= maxb; i++) ret = max(ret, (fw[i]));
  printf("%d\n", ret);
}
int main() {
  scanf("%d %d %d %d", &k, &n, &maxb, &t);
  if (t > n) t = n;
  if (t > maxb) t = maxb;
  for (int i = 1; i <= k; i++) solve();
  return 0;
}
