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 MAX_N = 100000;
const int MAX_S = 20000000;
const int INF = 0x3f3f3f3f;
pair<int, int> f[2][MAX_N + 10];
int p;
int n, maxb, pe;
int data[MAX_N + 10];
inline void readInt(int& a) {
  a = 0;
  int c;
  do c = getchar();
  while (c < '0' || c > '9');
  do {
    a = a * 10 - '0' + c;
    c = getchar();
  } while (c >= '0' && c <= '9');
}
int cnt[MAX_N + 10], order[MAX_N + 10];
void work() {
  for (int i = 1; i <= n; i++) readInt(data[i]);
  fill(cnt, cnt + maxb + 1, 0);
  for (int i = 1; i <= n; i++) ++cnt[data[i]];
  for (int i = 1; i <= maxb; i++) cnt[i] += cnt[i - 1];
  for (int i = 1; i <= n; i++) order[cnt[data[i]]--] = i;
  data[n + 1] = INF;
  f[0][0].first = f[0][0].second = 0;
  fill(f[0] + 1, f[0] + maxb + 1, make_pair(INF, n + 1));
  int cur = 0;
  pair<int, int> nxt;
  for (int i = 1; i <= n;) {
    for (int j = 0; j <= maxb; j++) f[cur ^ 1][j] = f[cur][j];
    do {
      int c = order[i];
      for (int j = maxb; j >= 0; j--) {
        if (data[f[cur][j].second] < data[c]) {
          nxt.second = c;
          nxt.first = f[cur][j].first + (f[cur][j].second > c);
          if (f[cur ^ 1][j + 1] > nxt) f[cur ^ 1][j + 1] = nxt;
        }
      }
      ++i;
    } while (i <= n && data[order[i]] == data[order[i - 1]]);
    cur ^= 1;
  }
  int ans;
  for (ans = maxb; f[cur][ans].first >= pe; ans--)
    ;
  printf("%d\n", ans);
}
int main() {
  int t;
  readInt(t);
  readInt(n);
  readInt(maxb);
  readInt(pe);
  while (t--) work();
  return 0;
}
