Problem:
Score : 200 points

Problem Statement

N Snukes called Snuke 1, Snuke 2, ..., Snuke N live in a town.

There are K kinds of snacks sold in this town, called Snack 1, Snack 2, ..., Snack K. The following d_i Snukes have Snack i: Snuke A_{i, 1}, A_{i, 2}, \cdots, A_{i, {d_i}}.

Takahashi will walk around this town and make mischief on the Snukes who have no snacks. How many Snukes will fall victim to Takahashi's mischief?

Constraints

All values in input are integers.

1 \leq N \leq 100

1 \leq K \leq 100

1 \leq d_i \leq N

1 \leq A_{i, 1} < \cdots < A_{i, d_i} \leq N

Input

Input is given from Standard Input in the following format:

N K
d_1
A_{1, 1} \cdots A_{1, d_1}
\vdots
d_K
A_{K, 1} \cdots A_{K, d_K}

Output

Print the answer.

Sample Input 1

3 2
2
1 3
1
3

Sample Output 1

1

Snuke 1 has Snack 1.

Snuke 2 has no snacks.

Snuke 3 has Snack 1 and 2.

Thus, there will be one victim: Snuke 2.

Sample Input 2

3 3
1
3
1
3
1
3

Sample Output 2

2

Solution in Java (OpenJDK 11.0.6) (judge verdict: Accepted)

import java.util.ArrayList;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		// TODO 自動生成されたメソッド・スタブ

		Scanner sc=new Scanner(System.in);
	  //すぬけ君の人数
		int N=sc.nextInt();

		//お菓子の種類数
		int K=sc.nextInt();

		ArrayList<Integer>array=new ArrayList<>();

		//お菓子の種類分繰り返し
		for(int i=0;i<K;i++) {
			//そのお菓子を持っている人数
			int	d=sc.nextInt();

			//そのお菓子を持っている人の名前の入力繰り返し
			for(int j=0;j<d;j++) {
				int checkint=sc.nextInt();
				if(!array.contains(checkint)) {
					//そのお菓子をもっている人の名前を配列に格納
					array.add(checkint);
				}
			}


		}
		System.out.println(N-array.size());

	}
}