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 Ruby (2.7.1) (judge verdict: Accepted)

require 'set'

def main
    n,k = gets.chomp.split(" ").map(&:to_i)
    ss = Array.new(n+1, 0)
    k.times do
        d = gets.chomp.to_i
        as = gets.split(" ").map(&:to_i)
        as.each do |i|
            ss[i] += 1
        end
    end
    puts ss.count{ |x| x == 0 } - 1
end

main

## イディオムなど
# until s.empty? do
#
## 配列のインデックス
# s[0,6] # 0番目から6文字
# s[0..6] # 0番目から6番目まで
# s[5..-1] # 5番目から最後まで
#
## 全探索
# [0, 1].repeated_permutation(n) do |combination| # bit全探索
# array.combination(i) do |selected| # array から要素が i 個である部分列を全部抽出
#
## 無向グラフの幅優先探索(BFS)
# to = Array.new(n).map{Array.new()} # ある頂点と繋がる頂点のリスト
# m.times do
#     a,b = gets.chomp.split(" ").map(&:to_i)
#     to[a].push b
#     to[b].push a
# end
# q = Queue.new # BFS は queue
# ds = Array.new(n, -1)
# pre = Array.new(n, -1)
#
## その他
# a1 = Marshal.load(Marshal.dump(a2)) # 深いコピー
# matrix = Array.new(n).map{Array.new()} # 二次元配列の初期化
#