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 T, n, m, k, f[101000], g[101000], ans, l, x;
vector<int> vec[101000];
int main() {
  for (cin >> T >> n >> m >> k; T--;) {
    k = min(k, min(n, m));
    l = n * k;
    for (int i = 0; i <= m; ++i) vec[i].clear();
    for (int i = 0; i < n; ++i) scanf("%d", &x), vec[x].push_back(i);
    memset(f, 33, sizeof f);
    f[0] = 0;
    for (int i = 1; i <= m; ++i)
      if (vec[i].size()) {
        memset(g, 33, sizeof g);
        for (auto o : vec[i]) {
          for (int j = 0; j < m && f[j] < l; ++j) {
            g[j] = min(g[j], f[j]);
            g[j + 1] = min(g[j + 1], f[j] % n <= o ? f[j] / n * n + o
                                                   : (f[j] / n + 1) * n + o);
          }
        }
        memcpy(f, g, sizeof(int) * (m + 2));
      }
    for (ans = 0; f[ans + 1] < l; ++ans)
      ;
    cout << ans << endl;
  }
}
