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 dp[100001], n, m, t, k, mb;
int ans, a[100001];
void r(int &x) {
  char g = getchar();
  while (g > '9' || g < '0') g = getchar();
  x = 0;
  while (g >= '0' && g <= '9') {
    x = x * 10 + g - '0';
    g = getchar();
  }
}
void solve() {
  memset(dp, 0, sizeof(dp));
  ans = 0;
  for (int i = 1; i <= n; i++) r(a[i]);
  if (t >= mb || t >= n) {
    for (int i = 1; i <= n; i++) {
      if (!dp[a[i]]) {
        dp[a[i]] = 1;
        ans++;
      }
    }
    printf("%d\n", ans);
    return;
  }
  int i, j, k;
  dp[mb] = 2e9;
  for (i = 1; i <= t; i++) {
    for (j = 1; j <= n; j++) {
      dp[a[j]] = dp[a[j] - 1] + 1;
      for (k = a[j] + 1; dp[k] < dp[k - 1]; dp[k] = dp[k - 1], k++)
        ;
    }
  }
  printf("%d\n", dp[mb - 1]);
}
int main() {
  r(k);
  r(n);
  r(mb);
  r(t);
  mb++;
  for (int i = 1; i <= k; i++) solve();
}
