Sereja and Sets

Sereja has m non-empty sets of integers A ₁, A ₂, …, A _m. What a lucky coincidence! The given sets are a partition of the set of all integers from 1 to n. In other words, for any integer v (1 ≤ v ≤ n) there is exactly one set A _t such that . Also Sereja has integer d.

Sereja decided to choose some sets from the sets he has. Let’s suppose that i ₁, i ₂, …, i _k (1 ≤ i ₁ < i ₂ < … < i _k ≤ m) are indexes of the chosen sets. Then let’s define an array of integers b, sorted in ascending order, as a union of the chosen sets, that is, . We’ll represent the element with number j in this array (in ascending order) as b _j. Sereja considers his choice of sets correct, if the following conditions are met:b₁ ≤ d; b_i + 1 - b_i ≤ d (1 ≤ i < |b|); n - d + 1 ≤ b_|b|.

Sereja wants to know what is the minimum number of sets (k) that he can choose so that his choice will be correct. Help him with that.

Input

The first line contains integers n, m, d (1 ≤ d ≤ n ≤ 10⁵, 1 ≤ m ≤ 20). The next m lines contain sets. The first number in the i-th line is s _i (1 ≤ s _i ≤ n). This number denotes the size of the i-th set. Then the line contains s _i distinct integers from 1 to n — set A _i.

It is guaranteed that the sets form partition of all integers from 1 to n.

Output

In a single line print the answer to the problem — the minimum value k at the right choice.

Examples

input

3 2 2
1 2
2 1 3

output

1

input

5 1 1
5 4 5 3 2 1

output

1

input

7 3 1
4 1 3 5 7
2 2 6
1 4

output

3

Solution:

#include <iostream>
#include <iomanip>
#include <cstdio>
#include <set>
#include <vector>
#include <map>
#include <cmath>
#include <algorithm>
#include <memory.h>
#include <string>
#include <cstring>
#include <sstream>
#include <cstdlib>
#include <ctime>
#include <cassert>

using namespace std;

const int N = 2000010;

int who[N], L[N], R[N], kb[N];
bool bad[N];

int main() {
int n, m, d;
scanf("%d %d %d", &n, &m, &d);
for (int i = 0; i < n; i++) who[i] = 0;
  for (int i = 0; i < m; i++) {
    int foo;
    scanf("%d", &foo);
    while (foo--) {
      int bar;
      scanf("%d", &bar);
      who[bar - 1] = (1 << i);
    }
  }
  int nn = (n + d - 1) / d * d;
  for (int i = n; i < nn; i++) who[i] = 0;
  int cur = 0;
  for (int i = 0; i < nn; i++) {
    if (i % d == 0) cur = 0;
    cur |= who[i];
    L[i] = cur;
  }
  cur = 0;
  for (int i = nn - 1; i >= 0; i--) {
if ((i + 1) % d == 0) cur = 0;
cur |= who[i];
R[i] = cur;
}
for (int i = 0; i < (1 << m); i++) bad[i] = false;
  for (int i = 0; i < n - d + 1; i++) {
    int now = R[i] | L[i + d - 1];
    bad[now] = true;
  }
  for (int bit = 0; bit < m; bit++)
    for (int i = 0; i < (1 << m); i++)
      if ((!(i & (1 << bit))) && bad[i]) bad[i | (1 << bit)] = true;
  kb[0] = 0;
  for (int i = 1; i < (1 << m); i++) kb[i] = kb[i & (i - 1)] + 1;
  int ans = 0;
  for (int i = 0; i < (1 << m); i++)
    if ((!bad[i]) && kb[i] > ans) ans = kb[i];
printf("%d\n", m - ans);
return 0;
}