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 = 2147483647;
const long long LLINF = 9223372036854775807LL;
const int maxn = 100010;
const long long cinf = 1000000010LL * 1000000010LL;
int a[maxn];
int main() {
  int k, n, maxb, t;
  scanf("%d%d%d%d", &k, &n, &maxb, &t);
  while (k--) {
    vector<vector<long long> > dp(n + 1, vector<long long>(maxb + 1, cinf));
    vector<vector<int> > nxt(n + 1, vector<int>(maxb + 1));
    for (int i = 0; i < n; ++i) scanf("%d", &a[i]), --a[i];
    for (int i = 0; i < maxb; ++i) nxt[n][i] = n;
    for (int i = n - 1; i >= 0; --i) {
      for (int j = 0; j < maxb; ++j) nxt[i][j] = nxt[i + 1][j];
      nxt[i][a[i]] = i;
    }
    for (int i = 0; i < maxb; ++i)
      if (nxt[0][i] != n) dp[1][i] = nxt[0][i];
    for (int i = 1; i < n; ++i)
      for (int j = 0; j < maxb - 1; ++j) {
        if (dp[i][j] == cinf) continue;
        dp[i][j + 1] = min(dp[i][j + 1], dp[i][j]);
        if (nxt[0][j + 1] == n) continue;
        int curpos = dp[i][j] % n;
        long long topos;
        if (nxt[curpos][j + 1] != n)
          topos = dp[i][j] - curpos + nxt[curpos][j + 1];
        else {
          topos = dp[i][j] + n - curpos + nxt[0][j + 1];
        }
        dp[i + 1][j + 1] = min(dp[i + 1][j + 1], topos);
      }
    int bestn = 0;
    long long mx = (long long)t * n - 1;
    for (int i = 0; i <= n; ++i)
      for (int j = 0; j < maxb; ++j)
        if (dp[i][j] <= mx) bestn = i;
    printf("%d\n", bestn);
  }
  return 0;
}
