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;
static const int INF = 500000000;
template <class T>
void debug(T a, T b) {
  for (; a != b; ++a) cerr << *a << ' ';
  cerr << endl;
}
int k, n, t2;
int b[100005];
int zip[100005];
int dp[100005], itr[100005];
void solve() {
  for (int i = 0; i < n; ++i) scanf("%d", &b[i]), zip[i] = b[i];
  sort(zip, zip + n);
  int zn = unique(zip, zip + n) - zip;
  for (int i = 0; i < n; ++i) b[i] = lower_bound(zip, zip + zn, b[i]) - zip;
  int T = min(t2, zn + 2);
  for (int i = 0; i < zn; ++i) dp[i] = INF;
  memset(itr, 0, sizeof(itr));
  for (int i = 0; i < T; ++i)
    for (int j = 0; j < n; ++j) {
      while (dp[itr[b[j]]] < b[j]) ++itr[b[j]];
      dp[itr[b[j]]] = b[j];
    }
  int res = 0;
  for (int i = 0; i < zn; ++i)
    if (dp[i] < INF) res = i + 1;
  printf("%d\n", res);
}
int main() {
  scanf("%d%d%*d%d", &k, &n, &t2);
  for (int hoge = 0; hoge < k; ++hoge) {
    solve();
  }
  return 0;
}
