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, c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = -1;
  for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
  return x * f;
}
const int MAXN = 100005;
int n, m, t, T, a[MAXN], l[MAXN], f[MAXN];
inline void Solve() {
  for (int i = 1; i <= n; i++) a[i] = l[i] = Read();
  sort(l + 1, l + n + 1), m = unique(l + 1, l + n + 1) - l - 1;
  if (t >= m) {
    printf("%d\n", m);
    return;
  }
  for (int i = 1; i <= n; i++) a[i] = lower_bound(l + 1, l + m + 1, a[i]) - l;
  for (int i = 1; i <= m; i++) f[i] = 0;
  for (int j = 1; j <= t; j++)
    for (int i = 1; i <= n; i++)
      for (int k = a[i]; k <= m; k++)
        if (f[k] < f[a[i] - 1] + 1)
          f[k] = f[a[i] - 1] + 1;
        else
          break;
  printf("%d\n", f[m]);
}
int main() {
  T = Read(), n = Read(), Read(), t = Read();
  while (T--) Solve();
  return 0;
}
