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 a[100010], d[100010], ans, dp[100010];
int lowbit(int x) { return x & -x; }
void update(int x, int v) {
  while (x <= maxb) {
    d[x] = max(d[x], v);
    x += lowbit(x);
  }
}
int query(int x) {
  int ret = 0;
  while (x) {
    ret = max(ret, d[x]);
    x -= lowbit(x);
  }
  return ret;
}
int main() {
  scanf("%d%d%d%d", &k, &n, &maxb, &t);
  for (int kase = 1; kase <= k; kase++) {
    ans = 0;
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
    if (t >= maxb) {
      sort(a + 1, a + n + 1);
      printf("%d\n", unique(a + 1, a + n + 1) - a - 1);
      continue;
    }
    memset(d, 0, sizeof d);
    memset(dp, 0, sizeof dp);
    int tmp;
    for (int j = 1; j <= t; j++)
      for (int i = 1; i <= n; i++) {
        tmp = query(a[i] - 1) + 1;
        if (tmp > dp[i]) {
          dp[i] = tmp;
          ans = max(ans, tmp);
          update(a[i], tmp);
        }
      }
    printf("%d\n", ans);
  }
}
