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 MX = 100005;
int a[MX];
vector<int> nxt[MX];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int tests, n, maxb, T;
  cin >> tests >> n >> maxb >> T;
  while (tests--) {
    for (int i = 0; i < n; i++) cin >> a[i];
    for (int i = 0; i < n; i++) a[i]--;
    nxt[n] = vector<int>(maxb, -1);
    for (int i = n - 1; i >= 0; i--) {
      nxt[i] = nxt[i + 1];
      nxt[i][a[i]] = i;
    }
    vector<long long> dp(n + 1, n * 1LL * T);
    dp[0] = 0;
    for (int x = 0; x < maxb; x++) {
      for (int len = n - 1; len >= 0; len--) {
        long long t;
        int pos = dp[len] % n;
        if (nxt[pos][x] == -1) {
          if (nxt[0][x] == -1)
            t = n * 1LL * T;
          else
            t = dp[len] - pos + n + nxt[0][x];
        } else {
          t = dp[len] - pos + nxt[pos][x];
        }
        dp[len + 1] = min(dp[len + 1], t);
      }
    }
    int ans = 0;
    for (int i = 0; i <= n; i++) {
      if (dp[i] < n * 1LL * T) ans = i;
    }
    cout << ans << "\n";
  }
}
