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, k, maxb, t;
int f[101001];
int l;
int a[101011];
void init() { cin >> k >> n >> maxb >> t; }
void force() {
  for (int T = 1; T <= k; T++) {
    for (int i = 1; i <= n; i++) cin >> a[i];
    for (int i = 1; i < t; i++) {
      for (int j = 1; j <= n; j++) a[i * n + j] = a[j];
    }
    int nt = n * t;
    memset(f, 0, sizeof(f));
    l = 0;
    for (int i = 1; i <= nt; i++)
      if (a[i] > f[l])
        f[++l] = a[i];
      else {
        int pos = upper_bound(f + 1, f + 1 + l, a[i]) - f;
        if (f[pos - 1] != a[i]) f[pos] = a[i];
      }
    cout << l << endl;
  }
}
int g[101010];
void work() {
  for (int T = 1; T <= k; T++) {
    for (int i = 1; i <= n; i++) cin >> a[i];
    memset(g, -1, sizeof(g));
    int lim = min(t, maxb);
    l = 0;
    for (int i = 1; i <= lim; i++) {
      for (int j = 1; j <= n; j++) {
        if (a[j] > f[l]) {
          f[++l] = a[j];
          g[a[j]] = l;
        } else {
          if (g[a[j]] == -1) {
            for (int p = a[j]; p <= maxb; p++)
              if (g[p] != -1) {
                f[g[p]] = a[j];
                g[a[j]] = g[p];
                g[p] = -1;
                break;
              }
          }
        }
      }
    }
    cout << l << endl;
  }
}
int main() {
  std::ios::sync_with_stdio(false);
  init();
  if (1LL * n * t <= 100000)
    force();
  else
    work();
  return 0;
}
