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 = 100005;
int k, n, maxb, t, fen[MAXN], ans[MAXN], b[MAXN], cnt[MAXN];
int get(int x) {
  int ret = 0;
  for (; x; x -= x & -x) ret = max(ret, fen[x]);
  return ret;
}
void upd(int x, int val) {
  for (; x < MAXN; x += x & -x) fen[x] = max(fen[x], val);
}
int solve() {
  memset(fen, 0, sizeof fen);
  memset(ans, 0, sizeof ans);
  memset(cnt, 0, sizeof cnt);
  int ret = 0, unique = 0, nx;
  for (int i = 0; i < n; ++i) unique += (cnt[b[i]]++ == 0);
  for (int i = 0; i < t && ret < unique; ++i)
    for (int j = 0; j < n; ++j) {
      nx = get(b[j] - 1) + 1;
      if (nx != ans[j]) {
        ans[j] = nx;
        ret = max(ret, ans[j]);
        upd(b[j], nx);
      }
    }
  return ret;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> k >> n >> maxb >> t;
  for (int i = 0; i < k; ++i) {
    for (int j = 0; j < n; ++j) cin >> b[j];
    cout << solve() << '\n';
  }
}
