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;
template <class T>
inline int size(const T& c) {
  return c.size();
}
const int maxn = 100000 + 10;
const int oo = 1000000000;
using namespace std;
pair<int, int> gt, g[maxn];
int a[maxn], k, t, maxb;
vector<int> b[maxn];
int vtmin[maxn], n;
bool operator<(pair<int, int> a, pair<int, int> b) {
  return a.first < b.first || (a.first == b.first && a.second < b.second);
}
pair<int, int> min(pair<int, int> a, pair<int, int> b) {
  if (a < b)
    return a;
  else
    return b;
}
void process() {
  for (int j = (1), _b = (n); j <= _b; j++) g[j] = make_pair(oo, 0);
  g[0] = make_pair(1, 0);
  for (int i = (1), _b = (maxb); i <= _b; i++) {
    if (b[i].size() == 0) continue;
    for (int j = (n), _a = (1); j >= _a; j--) {
      if (g[j - 1].first == oo) continue;
      vector<int>::iterator it =
          upper_bound(b[i].begin(), b[i].end(), g[j - 1].second);
      if (it != b[i].end()) {
        gt = make_pair(g[j - 1].first, *it);
      } else
        gt = make_pair(g[j - 1].first + 1, b[i][0]);
      g[j] = min(g[j], gt);
    }
  }
  int res = 1;
  for (int j = (1), _b = (n); j <= _b; j++)
    if (g[j].first <= t) res = j;
  cout << res << endl;
}
int main() {
  cin >> k >> n >> maxb >> t;
  for (int ss = (1), _b = (k); ss <= _b; ss++) {
    for (int i = (1), _b = (maxb); i <= _b; i++) b[i].clear();
    for (int i = (1), _b = (n); i <= _b; i++)
      scanf("%d", &a[i]), b[a[i]].push_back(i);
    for (int i = (1), _b = (maxb); i <= _b; i++) sort(b[i].begin(), b[i].end());
    process();
  }
  fclose(stdin);
  fclose(stdout);
  return 0;
}
