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;
const int maxn = 100005;
int k, n, maxb, t, b[maxn], m;
bool vis[maxn];
int id[maxn];
int Max[maxn];
int lowbit(int x) { return x & (-x); }
void update(int x, int v) {
  for (; x <= m; x += lowbit(x)) Max[x] = max(Max[x], v);
}
int query(int x) {
  int ans = 0;
  for (; x; x -= lowbit(x)) ans = max(ans, Max[x]);
  return ans;
}
int main() {
  scanf("%d%d%d%d", &k, &n, &maxb, &t);
  while (k--) {
    memset(vis, 0, sizeof(vis)), m = 0;
    for (int i = 1; i <= n; i++) scanf("%d", &b[i]), vis[b[i]] = 1;
    for (int i = 1; i <= maxb; i++)
      if (vis[i]) id[i] = ++m;
    if (m <= t) {
      printf("%d\n", m);
      continue;
    }
    for (int i = 1; i <= n; i++) b[i] = id[b[i]];
    memset(Max, 0, sizeof(Max));
    int ans = 0;
    for (int j = 1, res; j <= t; j++)
      for (int i = 1; i <= n; i++)
        res = query(b[i] - 1) + 1, update(b[i], res), ans = max(ans, res);
    printf("%d\n", ans);
  }
  return 0;
}
