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 INF = 1000000000;
int main() {
  ios_base::sync_with_stdio(0);
  int k, n, maxb, t;
  cin >> k >> n >> maxb >> t;
  for (int i = 0; i < k; ++i) {
    vector<int> b(n);
    for (int j = 0; j < n; ++j) {
      cin >> b[j];
      --b[j];
    }
    vector<vector<int> > nast(n, vector<int>(maxb, INF));
    nast[n - 1][b[n - 1]] = n - 1;
    for (int j = n - 2; j >= 0; --j) {
      for (int l = 0; l < maxb; ++l) nast[j][l] = nast[j + 1][l];
      nast[j][b[j]] = j;
    }
    vector<vector<int> > dp(n + 1, vector<int>(maxb, INF));
    for (int j = 0; j < maxb; ++j) dp[1][j] = nast[0][j];
    for (int j = 1; j < maxb; ++j) dp[1][j] = min(dp[1][j], dp[1][j - 1]);
    for (int j = 2; j <= n; ++j) {
      for (int l = 1; l < maxb; ++l) {
        int ost = dp[j - 1][l - 1];
        if (ost == INF) {
          dp[j][l] = dp[j][l - 1];
          continue;
        }
        int kol = nast[ost % n][l];
        if (kol != INF)
          dp[j][l] = ost - (ost % n) + kol;
        else {
          kol = nast[0][l];
          if (kol != INF) dp[j][l] = ost - (ost % n) + n + kol;
        }
        dp[j][l] = min(dp[j][l - 1], dp[j][l]);
      }
    }
    int odp;
    for (int j = 1; j <= n; ++j)
      if (dp[j][maxb - 1] != INF && dp[j][maxb - 1] < (long long)n * t) odp = j;
    cout << odp << endl;
  }
  return 0;
}
