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 k, n, maxb, t;
int b[100005], c[100005], f[100005];
char getc() {
  static char *S, *T, buf[65536];
  if (S == T) {
    T = (S = buf) + fread(buf, 1, 65536, stdin);
    if (S == T) return EOF;
  }
  return *S++;
}
int read() {
  static char ch;
  static int D;
  while (!isdigit(ch = getc()))
    ;
  for (D = ch - '0'; isdigit(ch = getc());) D = D * 10 + ch - '0';
  return D;
}
void add(int x, int y) {
  for (; x <= maxb; x += (x & (-x))) c[x] = max(c[x], y);
}
int query(int x) {
  int ret = 0;
  for (; x; x -= (x & (-x))) ret = max(ret, c[x]);
  return ret;
}
int main() {
  k = read();
  n = read();
  maxb = read();
  t = read();
  for (int tim = 1; tim <= k; tim++) {
    for (int i = 1; i <= n; i++) b[i] = read();
    memset(f, 0, sizeof(f));
    int x = min(t, maxb) * n;
    for (int i = 1; i <= min(t, maxb); i++) {
      for (int j = 1; j <= n; j++) {
        for (int k = b[j]; k <= maxb; k++) {
          if (f[b[j] - 1] + 1 > f[k])
            f[k] = f[b[j] - 1] + 1;
          else
            break;
        }
      }
    }
    int ans = 0;
    for (int i = 1; i <= maxb; i++) ans = max(ans, f[i]);
    printf("%d\n", ans);
  }
}
