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 DE = 0;
const int Maxn = 200005;
int n, K, m, T;
pair<int, int> dp[2][Maxn];
int a[Maxn], b[Maxn];
int main() {
  int i, j;
  while (~scanf("%d%d%d%d", &K, &n, &m, &T))
    while (K--) {
      for (i = 1; i <= n; i++) scanf("%d", &a[i]);
      for (i = 1; i <= n; i++) b[i] = a[i];
      sort(b + 1, b + n + 1);
      m = unique(b + 1, b + n + 1) - b - 1;
      for (i = 1; i <= n; i++) a[i] = lower_bound(b + 1, b + m + 1, a[i]) - b;
      for (i = 1; i <= m; i++) dp[0][i] = make_pair(T + 1, 1);
      dp[0][0] = make_pair(0, 0);
      for (i = 1; i <= m; i++) {
        int t = i & 1;
        for (j = 1; j <= m; j++)
          dp[t ^ 1][j] = min(dp[t ^ 1][j], dp[t ^ 1][j - 1]);
        for (j = 0; j <= m; j++) dp[t][j] = make_pair(T + 1, 1);
        for (j = 1; j <= n; j++) {
          pair<int, int> th = dp[t ^ 1][a[j] - 1];
          if (th.second >= j) th.first++;
          th.second = j;
          dp[t][a[j]] = min(dp[t][a[j]], th);
        }
        if (DE) {
          printf("i=%d\t", i);
          for (j = 1; j <= m; j++)
            printf("(%d,%d) ", dp[t][j].first, dp[t][j].second);
          printf("\n");
        }
        for (j = 1; j <= m; j++)
          if (dp[t][j].first < T) break;
        if (j > m) break;
      }
      printf("%d\n", i - 1);
    }
  return 0;
}
