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 = 1e5 + 9;
const long long mod = 1e9 + 7;
int maxb;
int b[maxn];
int fen[maxn];
bool vis[maxn];
void clearfen() { memset(fen, 0, sizeof fen); }
void upd(int x, int val) {
  for (; x < maxb; x += x & -x) fen[x] = max(fen[x], val);
}
int get(int x) {
  int res = 0;
  if (x == 0) return 0;
  for (; x; x -= x & -x) res = max(res, fen[x]);
  return res;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int k, n, t;
  cin >> k >> n >> maxb >> t;
  maxb += 10;
  t = min(n, t);
  for (int i = 0; i < k; i++) {
    clearfen();
    memset(vis, 0, sizeof vis);
    int ans = 0;
    for (int j = 0; j < n; j++) {
      cin >> b[j];
      if (!vis[b[j]]) ans++;
      vis[b[j]] = true;
    }
    if (t > maxb) {
      cout << ans << "\n";
      continue;
    }
    ans = 0;
    for (int x = 0; x < t; x++)
      for (int j = 0; j < n; j++) upd(b[j], get(b[j] - 1) + 1);
    for (int j = 0; j < n; j++) ans = max(ans, get(b[j]));
    cout << ans << "\n";
  }
}
