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;
int nq, n, m, t, tot, ans;
int a[100010], sa[100010], g[100010], f[20000010];
int main() {
  int i, j, q;
  scanf("%d%d%d%d", &nq, &n, &m, &t);
  for (q = 1; q <= nq; q++) {
    for (i = 1; i <= n; i++) {
      scanf("%d", &a[i]);
      sa[i] = a[i];
    }
    sort(sa + 1, sa + n + 1);
    tot = 0;
    for (i = 1; i <= n; i++)
      if (sa[i] != sa[i - 1]) sa[++tot] = sa[i];
    memset(f, 0, sizeof(f));
    memset(g, 0, sizeof(g));
    ans = 0;
    for (i = 1; i <= (long long)n * t && i <= n * m; i++) {
      f[i] = g[a[(i - 1) % n + 1] - 1] + 1;
      ans = max(ans, f[i]);
      j = a[(i - 1) % n + 1];
      while (j <= m && g[j] < f[i]) {
        g[j] = f[i];
        j++;
      }
    }
    printf("%d\n", ans);
  }
  return 0;
}
