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;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch > '9' || ch < '0') ch == '-' && (f = 0) || (ch = getchar());
  while (ch <= '9' && ch >= '0')
    x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
  return f ? x : -x;
}
int b[100020], f[100020];
int k = read(), n = read(), maxb = read(), t = read();
void update(int x, int v) {
  for (; x <= maxb; x += x & -x) f[x] = max(f[x], v);
}
int query(int x) {
  int res = 0;
  for (; x; x ^= x & -x) res = max(res, f[x]);
  return res;
}
int ans[100020];
int main(int argc, char const *argv[]) {
  t = min(min(n, maxb), t);
  while (k--) {
    for (int i = 1; i <= n; i++) b[i] = read();
    memset(f, 0, sizeof f);
    memset(ans, 0, sizeof ans);
    for (int j = 1; j <= t; j++) {
      for (int i = 1; i <= n; i++) {
        int res = query(b[i] - 1) + 1;
        if (ans[b[i]] < res) {
          ans[b[i]] = res;
          update(b[i], res);
        }
      }
    }
    cout << query(maxb) << endl;
  }
  return 0;
}
