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 N = 100100;
int k, n, maxn, t, sum;
int a[N], v[200 * N];
int f[N], tree[N];
int lowbit(int x) { return x & (-x); }
void update(int x, int y) {
  for (; x <= maxn; x += lowbit(x)) tree[x] = max(tree[x], y);
}
int find(int x) {
  int ans = 0;
  for (; x; x -= lowbit(x)) ans = max(ans, tree[x]);
  return ans;
}
int main() {
  scanf("%d%d%d%d", &k, &n, &maxn, &t);
  while (k--) {
    sum = 0;
    for (int i = 1; i <= n; i++) {
      scanf("%d", &a[i]);
      if (v[a[i]] != k + 1) sum++;
      v[a[i]] = k + 1;
    }
    if (t >= sum) {
      printf("%d\n", sum);
      continue;
    }
    int ans = 0;
    for (int i = 1; i <= N; i++) f[i] = tree[i] = 0;
    for (int i = 1; i <= t; i++)
      for (int j = 1; j <= n; j++) {
        int x = find(a[j] - 1) + 1;
        if (x > f[j]) {
          f[j] = x;
          ans = max(ans, x);
          update(a[j], x);
        }
        if (ans >= sum) break;
      }
    printf("%d\n", ans);
  }
  return 0;
}
