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;
void readn(int &x) {
  char c;
  x = 0;
  do {
    c = getchar();
  } while (c < '0' || c > '9');
  do {
    x = x * 10 + c - '0';
    c = getchar();
  } while (c <= '9' && c >= '0');
}
const int MAXN = 101 * 1000;
int k, n, maxb, t;
int a[MAXN], fen[MAXN], v[MAXN];
inline void add(int pos, int val) {
  v[pos] = val;
  while (pos <= maxb) {
    if (fen[pos] < val) fen[pos] = val;
    pos += pos & -pos;
  }
}
inline int rmax(int pos) {
  int tot = 0;
  while (pos) {
    if (tot < fen[pos]) tot = fen[pos];
    pos -= pos & -pos;
  }
  return tot;
}
int solve() {
  memset(fen, 0, sizeof fen);
  memset(v, 0, sizeof v);
  for (register int rep = (0); rep < (int)(t); ++rep)
    for (register int i = (0); i < (int)(n); ++i)
      if (a[i]) {
        add(a[i], rmax(a[i] - 1) + 1);
        if (rep + 1 >= a[i] || v[a[i]] == a[i]) a[i] = 0;
      }
  return rmax(maxb);
}
int main() {
  scanf("%d %d %d %d", &k, &n, &maxb, &t);
  t = min(n, min(t, maxb));
  while (k--) {
    for (register int i = (0); i < (int)(n); ++i) readn(a[i]);
    printf("%d\n", solve());
  }
  return 0;
}
