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, m, T, i, j, L, wz, dp[100005], ans, MIN, DP[100005];
struct node {
  int x, y;
} a[100005];
int cmp(node i, node j) { return i.x < j.x; }
int main() {
  scanf("%d%d%d%d", &k, &n, &m, &T);
  while (k--) {
    for (i = 1; i <= n; i++) scanf("%d", &a[i].x);
    for (i = 1; i <= n; i++) a[i].y = i;
    sort(a + 1, a + n + 1, cmp);
    for (i = 1; i <= n; i++) dp[i] = 1;
    ans = 1;
    for (i = 2; i <= m; i++) {
      L = 1;
      MIN = T + 1;
      for (j = 1; j <= n; j++) {
        DP[j] = MIN + (wz > a[j].y);
        if (DP[j] <= T) ans = i;
        if (a[j].x != a[j + 1].x) {
          while (L <= j) {
            if (dp[L] < MIN) {
              MIN = dp[L];
              wz = a[L].y;
            } else if (dp[L] == MIN)
              wz = min(wz, a[L].y);
            L++;
          }
        }
      }
      for (j = 1; j <= n; j++) dp[j] = DP[j];
    }
    cout << ans << endl;
  }
  return 0;
}
