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 std::cin;
using std::cout;
using std::endl;
long long Factorial(long long i) { return i > 1 ? i * Factorial(i - 1) : 1; }
long long Combination(long long n, long long r) {
  if (n - r < r) r = n - r;
  long long ansn = 1, ansr = 1;
  for (long long i = n - r + 1; i <= n; ++i) ansn *= i;
  for (long long j = 1; j <= r; ++j) ansr *= j;
  return ansn / ansr;
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
std::vector<int> Eratosthenes(const int len, int *vis) {
  memset(vis, 0, sizeof(int) * len);
  std::vector<int> prime;
  int m = (int)sqrt(len + 0.5);
  for (int i = (2); i <= (m); ++i)
    if (!vis[i])
      for (int j = i * i; j < len; j += i) vis[j] = 1;
  for (int i = (2); i < (len); ++i)
    if (!vis[i]) prime.push_back(i);
  return prime;
}
void phi_table(int n, int *phi) {
  for (int i = (2); i <= (n); ++i) phi[i] = 0;
  phi[1] = 1;
  for (int i = (2); i <= (n); ++i)
    if (!phi[i])
      for (int j = i; j <= n; j += i) {
        if (!phi[j]) phi[j] = j;
        phi[j] = phi[j] / i * (i - 1);
      }
}
const int maxn = 1e5 + 10;
int in[maxn], f[maxn];
int k, n, b, t;
int main() {
  scanf("%d%d%d%d", &k, &n, &b, &t);
  t = std::min(t, std::min(b, n));
  while (k--) {
    for (int i = (1); i <= (n); ++i) scanf("%d", in + i);
    memset(f, 0, sizeof(f));
    for (int i = (1); i <= (t); ++i)
      for (int j = (1); j <= (n); ++j) {
        f[in[j]] = f[in[j] - 1] + 1;
        for (int k = in[j] + 1; k <= b && f[in[j]] > f[k]; ++k) f[k] = f[in[j]];
      }
    printf("%d\n", f[b]);
  }
  return 0;
}
