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;
int TC, N, mxb, t, Xn;
int B[100010];
vector<int> X;
map<int, int> dx;
vector<vector<int> > prec;
int la[5010];
pair<int, int> dp[2][5010];
int main() {
  scanf("%d %d %d %d", &TC, &N, &mxb, &t);
  while (TC--) {
    X.clear();
    dx.clear();
    for (int i = 0; i < N; i++) {
      scanf("%d", &B[i]);
      X.push_back(B[i]);
    }
    sort(X.begin(), X.end());
    X.resize(unique(X.begin(), X.end()) - X.begin());
    Xn = X.size();
    for (int i = 0; i < Xn; i++) dx[X[i]] = i;
    for (int i = 0; i < N; i++) {
      B[i] = dx[B[i]];
    }
    prec = vector<vector<int> >(N, vector<int>(Xn));
    memset(la, -1, sizeof(la));
    for (int i = 2 * N - 1; i >= 0; i--) {
      for (int j = 0; j < Xn; j++) {
        prec[i % N][j] = la[j];
      }
      la[B[i % N]] = i % N;
    }
    int ans = 0;
    for (int i = 0; i < 2; i++)
      for (int j = 0; j <= Xn; j++) dp[i][j] = pair<int, int>(1e9, 1e9);
    dp[0][0] = pair<int, int>(1, -1);
    for (int i = 0; i < Xn; i++) {
      for (int j = 0; j <= Xn; j++) {
        pair<int, int> p = dp[i & 1][j];
        if (p.first > t) continue;
        dp[(i + 1) & 1][j + 1] = min(dp[(i + 1) & 1][j + 1], p);
        dp[(i + 1) & 1][j] =
            min(dp[(i + 1) & 1][j],
                prec[(N + p.second) % N][i] <= p.second
                    ? pair<int, int>(p.first + 1, prec[(N + p.second) % N][i])
                    : pair<int, int>(p.first, prec[(N + p.second) % N][i]));
      }
      for (int j = 0; j <= Xn; j++) {
        dp[i & 1][j] = pair<int, int>(1e9, 1e9);
      }
    }
    for (int i = 0; i <= Xn; i++) {
      if (dp[Xn & 1][i].first <= t) {
        ans = Xn - i;
        break;
      }
    }
    printf("%d\n", ans);
  }
}
