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 long double pi = 3.14159265359;
template <typename T>
T abs(T x) {
  return x > 0 ? x : -x;
}
template <typename T>
T sqr(T x) {
  return x * x;
}
template <typename T>
void chmin(T &x, T y) {
  x = min(x, y);
}
template <typename T>
void chmax(T &x, T y) {
  x = max(x, y);
}
int main() {
  int k, n, bm, t;
  scanf("%d %d %d %d", &k, &n, &bm, &t);
  while (k--) {
    vector<int> cc;
    vector<int> a;
    for (int i = 0; i < n; i++) {
      int x;
      scanf("%d", &x);
      a.push_back(x);
      cc.push_back(x);
    }
    sort(cc.begin(), cc.end());
    cc.resize(unique(cc.begin(), cc.end()) - cc.begin());
    for (int &x : a) {
      x = lower_bound(cc.begin(), cc.end(), x) - cc.begin();
    }
    int ub = (int)cc.size() + 2;
    vector<int> f(ub, 0);
    for (int i = 0; i < min(t, ub); i++) {
      for (int x : a) {
        int y = f[x] + 1;
        for (int j = x + 1; j < ub; j++) {
          if (f[j] >= y) {
            break;
          }
          f[j] = y;
        }
      }
    }
    cout << *max_element(f.begin(), f.end()) << endl;
  }
  return 0;
}
