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, maxb, t;
int a[200001], S[200001];
int ans[200001];
bool vis[200001];
int maxl;
void add(int x, int v) {
  for (int i = x; i <= maxb; i += (i & (-i))) S[i] = max(S[i], v);
}
int sum(int x) {
  int res = 0;
  for (int i = x; i; i -= (i & (-i))) {
    res = max(res, S[i]);
  }
  return res;
}
int main() {
  scanf("%d", &k);
  scanf("%d", &n);
  scanf("%d", &maxb);
  scanf("%d", &t);
  for (int i = 1; i <= k; i++) {
    for (int j = 1; j <= n; j++) scanf("%d", &a[j]);
    maxl = 0;
    if (t >= maxb) {
      memset(vis, 0, sizeof(vis));
      for (int p = 1; p <= n; p++) {
        if (!vis[a[p]]) {
          maxl++;
          vis[a[p]] = 1;
        }
      }
      printf("%d\n", maxl);
      continue;
    }
    memset(S, 0, sizeof(S));
    memset(ans, 0, sizeof(ans));
    for (int r = 1; r <= t; r++) {
      for (int l = 1; l <= n; l++) {
        int it = sum(a[l] - 1) + 1;
        int hh = ans[a[l]];
        ans[a[l]] = max(ans[a[l]], it);
        if (ans[a[l]] != hh) add(a[l], it);
      }
    }
    maxl = -1;
    for (int j = 1; j <= maxb; j++) maxl = max(maxl, ans[j]);
    printf("%d\n", maxl);
    continue;
  }
}
