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>
const int MAXN = 100005;
const int MAXS = 100005;
int n, maxb, t;
int b[MAXN];
int f[MAXS];
int s[MAXS];
void init() {
  for (int i = 0; i < n; ++i) {
    scanf("%d", &b[i]);
    --b[i];
  }
}
void insert(int x, int c) {
  ++x;
  while (x <= maxb) {
    if (c > s[x]) s[x] = c;
    x += x & -x;
  }
}
int query(int x) {
  int max = 0;
  while (x) {
    if (s[x] > max) max = s[x];
    x -= x & -x;
  }
  return max;
}
void solve() {
  if (t > maxb) {
    std::sort(b, b + n);
    int cnt = std::unique(b, b + n) - b;
    printf("%d\n", cnt);
    return;
  }
  memset(f, 0, sizeof(f));
  memset(s, 0, sizeof(f));
  for (int i = 0; i < t; ++i)
    for (int j = 0; j < n; ++j) {
      int k = query(b[j]) + 1;
      if (k > f[j]) {
        f[j] = k;
        insert(b[j], k);
      }
    }
  int ans = 0;
  for (int i = 0; i < n; ++i)
    if (f[i] > ans) ans = f[i];
  printf("%d\n", ans);
}
int main() {
  int k;
  scanf("%d%d%d%d", &k, &n, &maxb, &t);
  while (k--) {
    init();
    solve();
  }
  return 0;
}
