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 = 100000 + 10;
const long long INF = 1000000000000LL;
int n, k, maxb;
long long t;
int b[N];
long long f[2][N];
vector<int> app[N];
inline int getNxt(const vector<int> &vec, int cur) {
  int p = upper_bound(vec.begin(), vec.end(), cur) - vec.begin();
  if (p == vec.size()) {
    return n - cur + vec[0];
  } else {
    return vec[p] - cur;
  }
}
void solve() {
  for (int i = 0; i <= maxb + 1; ++i) {
    app[i].clear();
  }
  for (int i = 0; i < n; ++i) {
    scanf("%d", b + i);
    app[b[i]].push_back(i);
  }
  for (int i = 0; i <= maxb; ++i) {
    f[0][i] = INF;
  }
  for (int i = 0; i < n; ++i) {
    f[0][b[i]] = min(f[0][b[i]], (long long)i);
  }
  for (int j = 1; j <= maxb; ++j) {
    f[0][j] = min(f[0][j], f[0][j - 1]);
  }
  int ret = 1;
  for (int i = 1; i <= n; ++i) {
    int nxt = i & 1;
    int cur = nxt ^ 1;
    for (int j = 0; j <= maxb; ++j) {
      f[nxt][j] = INF;
    }
    for (int j = 0; j <= maxb; ++j) {
      long long p = f[cur][j];
      if (p >= INF) continue;
      ret = i;
      if (app[j + 1].size() == 0) continue;
      int x = p % n;
      int y = getNxt(app[j + 1], x);
      if (p + y >= t * n) {
        continue;
      }
      f[nxt][j + 1] = p + y;
    }
    if (ret < i) break;
    for (int j = 1; j <= maxb; ++j) {
      f[nxt][j] = min(f[nxt][j], f[nxt][j - 1]);
    }
  }
  cout << ret << endl;
}
int main() {
  cin >> k >> n >> maxb >> t;
  for (int i = 0; i < k; ++i) {
    solve();
  }
  return 0;
}
