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 n, vA, m, A[100010];
inline void Read(int &digit) {
  digit = 0;
  char c;
  for (c = getchar(); c < '0' || c > '9'; c = getchar())
    ;
  for (; c >= '0' && c <= '9'; digit = digit * 10 + c - '0', c = getchar())
    ;
}
void read() {
  for (int i = 1; i <= n; ++i) Read(A[i]);
}
int LIS() {
  static bool use[100010];
  static int dp[100010], tmp[100010];
  static int b[100010];
  int cnt = 0;
  for (int i = 1; i <= n; ++i) use[A[i]] = true;
  for (int i = 1; i <= vA; ++i)
    if (use[i]) b[++cnt] = i, use[i] = false;
  if (cnt <= m) return cnt;
  memset(dp, 0, sizeof(dp));
  for (int i = 1; i <= cnt; ++i) {
    int minv = dp[0];
    bool mark = true;
    dp[0] = 1 << 30;
    for (int j = 1; j <= cnt; ++j) {
      int x = b[j];
      tmp[x] = minv;
      if (minv > dp[x]) minv = dp[x];
      dp[x] = 1 << 30;
    }
    for (int j = n; j >= 1; --j) {
      int v = A[j], x = (tmp[v] - 1 + n) % n + 1,
          val = (tmp[v] - 1 + n) / n * n + j;
      if (x < j) val -= n;
      if (dp[v] > val) dp[v] = val;
      if (dp[v] <= n * m) mark = false;
    }
    if (mark) return i - 1;
  }
  return cnt;
}
int main() {
  int T;
  for (Read(T), Read(n), Read(vA), Read(m); T; --T) {
    read();
    printf("%d\n", LIS());
  }
  return 0;
}
