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 INT_INF = (1U << 31) - 10;
const int MAX_B = 100010, MAX_N = 100010;
int b[MAX_N];
int dp[MAX_N];
set<int> s;
int main(int argc, char *argv[]) {
  int K, N, maxb, T;
  cin >> K >> N >> maxb >> T;
  for (int k = 0; k < K; k++) {
    s.clear();
    for (int i = 0; i < N; i++) {
      scanf("%d", &(b[i]));
      s.insert(b[i]);
    }
    if (T >= s.size()) {
      printf("%ld\n", s.size());
      continue;
    }
    fill(dp, dp + MAX_N, INT_INF);
    for (int i = 0; i < T; i++) {
      for (int j = 0; j < N; j++) {
        if (b[j] < i) continue;
        *lower_bound(dp, dp + N, b[j]) = b[j];
      }
    }
    printf("%ld\n", lower_bound(dp, dp + N, INT_INF) - dp);
  }
  return 0;
}
