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>
#pragma comment(linker, "/stack:256000000")
using namespace std;
const long long INF = (long long)1e18;
vector<long long> dp[100100];
vector<int> nxt[100100];
int a[200200];
int lst[100100];
int main() {
  int k, n, mx, t;
  scanf("%d %d %d %d", &k, &n, &mx, &t);
  long long L = (long long)n * t;
  for (int(i) = 0; (i) < (n + 1); (i)++) dp[i].resize(mx + 1);
  for (int(i) = 0; (i) < (n + 1); (i)++) nxt[i].resize(mx + 1);
  for (int(_) = 0; (_) < (k); (_)++) {
    for (int(i) = 0; (i) < (n); (i)++) scanf("%d", &a[i]);
    memcpy(a + n, a, 4 * n);
    for (int(i) = 0; (i) < (n + 1); (i)++)
      for (int(j) = 0; (j) < (mx + 1); (j)++) nxt[i][j] = -1;
    memset(lst, -1, sizeof(lst));
    for (int i = 2 * n - 1; i >= n; i--) {
      lst[a[i]] = i;
    }
    for (int i = n - 1; i >= 0; i--) {
      for (int j = 1; j <= mx; j++) {
        nxt[i][j] = lst[j];
      }
      lst[a[i]] = i;
    }
    for (int(i) = 0; (i) < (n + 1); (i)++)
      for (int(j) = 0; (j) < (mx + 1); (j)++) dp[i][j] = INF;
    for (int i = n - 1; i >= 0; i--) dp[1][a[i]] = i + 1;
    for (int i = 1; i < n; i++) {
      for (int j = 1; j <= mx; j++) {
        if (j == mx) continue;
        if (dp[i][j] == INF) continue;
        dp[i][j + 1] = min(dp[i][j + 1], dp[i][j]);
        int cur = (dp[i][j] - 1) % n;
        int to = nxt[cur][j + 1];
        if (to == -1) continue;
        int step = to - cur;
        if (dp[i][j] + step > L) continue;
        dp[i + 1][j + 1] = min(dp[i + 1][j + 1], dp[i][j] + step);
      }
    }
    int res = -1;
    for (int i = n; i >= 1; i--) {
      for (int(j) = 0; (j) < (mx + 1); (j)++) {
        if (dp[i][j] < INF) {
          res = i;
          break;
        }
      }
      if (res != -1) break;
    }
    printf("%d\n", res);
  }
  return 0;
}
