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 C++ (GCC 9.2.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
#define boost ios::sync_with_stdio(false); cin.tie(0)
#define ll              long long int
#define ld              long double
#define mk              make_pair
#define pb              push_back
#define f               first
#define s               second
#define fo(i,a,b)       for(i=a;i<b;i++)
#define foe(i,a,b)      for(i=a;i<=b;i++)
#define all(x)          x.begin(),x.end()
#define vi              vector<int>
#define vl              vector <long long int>
#define pii             pair <int,int>
#define pll             pair <long long int, long long int>
#define vpii            vector< pair<int,int> >
#define vpll            vector < pair <long long int,long long int> >
#define MOD             1000000007
using namespace std;
const int inf = INT_MAX;
const ll inf64 = LLONG_MAX;


int main()
{
    boost;
    ll n,k;
    cin >> n >> k;
    ll i;
    vi cnt(n);
    fo(i,0,k)
    {
    	ll d;
    	cin >> d;
    	while(d--)
    	{
    		ll num;
    		cin >> num;
    		num--;
    		cnt[num]++;
    	}
    }
    ll ans = 0;
    fo(i,0,n)if(cnt[i]==0)ans++;
    cout << ans;
    
    return 0;
}