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 N = 1e5 + 9;
const int M = 2e7 + 9;
int n, t, m;
int dp[M], b[N], ma[M];
void add(int i, int v) {
  for (; i <= m; i += i & -i) {
    if (ma[i] >= v) break;
    ma[i] = max(ma[i], v);
  }
}
int ask(int i, int rex = 0) {
  for (; i; i -= i & -i) rex = max(rex, ma[i]);
  return rex;
}
int read() {
  int rex = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') ch = getchar();
  while (ch >= '0' && ch <= '9') rex = rex * 10 + ch - '0', ch = getchar();
  return rex;
}
void work() {
  int ans = 0;
  for (int i = 1; i <= n; ++i) b[i] = read();
  for (int j = 1; j <= t; ++j) {
    for (int i = 1; i <= n; ++i) {
      int y = ask(b[i] - 1);
      if (y + 1 <= b[i]) add(b[i], y + 1);
      ans = max(ans, y + 1);
      if (ans == m) break;
    }
  }
  for (int i = 1; i <= m; ++i) ma[i] = 0;
  printf("%d\n", ans);
}
int main() {
  int T;
  scanf("%d%d%d%d", &T, &n, &m, &t);
  t = min(t, m);
  while (T--) work();
  return 0;
}
