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 INF = ((long long)1E9 + 7);
template <class C>
void mini(C& a, C b) {
  a = min(a, b);
}
template <class C>
void maxi(C& a, C b) {
  a = max(a, b);
}
const int MAXN = 100100;
int main() {
  ios_base::sync_with_stdio(false);
  int k, n, mb, t;
  cin >> k >> n >> mb >> t;
  t = min(t, n + 10);
  for (int _ = 0; _ < (k); ++_) {
    vector<int> b(n, 0);
    map<int, int> s;
    for (int i = 0; i < (n); ++i) {
      cin >> b[i];
      s[b[i]] = 0;
    }
    int l = 0;
    for (__typeof((s).begin()) it = ((s).begin()); it != (s).end(); ++it) {
      it->second = l;
      ++l;
    }
    for (int i = 0; i < (n); ++i) b[i] = s[b[i]];
    int m = ((int)(s).size());
    vector<vector<int> > next(n);
    vector<int> pr(m);
    for (int i = (n - 1); i >= (0); --i) {
      pr[b[i]] = i;
    }
    for (int i = (n - 1); i >= (0); --i) {
      next[i] = pr;
      pr[b[i]] = i;
    }
    vector<vector<long long> > tab(m, vector<long long>(m, INF));
    tab[0][0] = pr[0];
    for (int i = (1); i <= (m - 1); ++i) {
      tab[0][i] = min((long long)pr[i], tab[0][i - 1]);
    }
    int best = 1;
    for (int i = (1); i <= (m); ++i) {
      for (int j = (i); j <= (m - 1); ++j) {
        long long p = tab[i - 1][j - 1];
        long long nx = next[p % n][j];
        if (nx < p % n) nx += n;
        nx += n * (p / n);
        tab[i][j] = min(nx, tab[i][j - 1]);
        if (tab[i][j] < (long long)t * n) best = i + 1;
      }
    }
    cout << best << endl;
  }
  return 0;
}
