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 Mod = 1e9 + 7;
inline int FIX(long long a) { return (a % Mod + Mod) % Mod; }
int k, n, maxb, t;
int limit;
int b[100000 + 5];
int a[100000 + 5];
void add(int p, int v) {
  p = maxb - p + 1;
  for (int i = p; i <= maxb; i += (i & -i)) {
    a[i] = max(a[i], v);
  }
}
int query(int p) {
  p = maxb - p + 1;
  int ret = 0;
  for (int i = p; i >= 1; i -= (i & -i)) {
    ret = max(ret, a[i]);
  }
  return ret;
}
int ans[100000 + 5];
int main() {
  cin >> k >> n >> maxb >> t;
  while (k--) {
    for (int i = 1; i <= n; i++) {
      scanf("%d", &b[i]);
    }
    limit = min(t - 1, maxb) + 2;
    int M = min(t - 1, maxb);
    memset(a, 0, sizeof(a));
    memset(ans, 0, sizeof(ans));
    for (int j = 0; j <= M; j++) {
      for (int i = n; i >= 1; i--) {
        int cur = query(b[i] + 1) + 1;
        if (cur > ans[i]) {
          ans[i] = cur;
          add(b[i], cur);
        }
      }
    }
    cout << *max_element(ans + 1, ans + n + 1) << endl;
  }
  return 0;
}
