The "BerCorp" company has got n employees. These employees can use m approved official languages for the formal correspondence. The languages are numbered with integers from 1 to m. For each employee we have the list of languages, which he knows. This list could be empty, i. e. an employee may know no official languages. But the employees are willing to learn any number of official languages, as long as the company pays their lessons. A study course in one language for one employee costs 1 berdollar.

Find the minimum sum of money the company needs to spend so as any employee could correspond to any other one (their correspondence can be indirect, i. e. other employees can help out translating).

Input

The first line contains two integers n and m (2 ≤ n, m ≤ 100) — the number of employees and the number of languages.

Then n lines follow — each employee's language list. At the beginning of the i-th line is integer ki (0 ≤ ki ≤ m) — the number of languages the i-th employee knows. Next, the i-th line contains ki integers — aij (1 ≤ aij ≤ m) — the identifiers of languages the i-th employee knows. It is guaranteed that all the identifiers in one list are distinct. Note that an employee may know zero languages.

The numbers in the lines are separated by single spaces.

Output

Print a single integer — the minimum amount of money to pay so that in the end every employee could write a letter to every other one (other employees can help out translating).

Examples

Input

5 5
1 2
2 2 3
2 3 4
2 4 5
1 5


Output

0


Input

8 7
0
3 1 2 3
1 1
2 5 4
2 6 7
1 3
2 7 4
1 1


Output

2


Input

2 2
1 2
0


Output

1

Note

In the second sample the employee 1 can learn language 2, and employee 8 can learn language 4.

In the third sample employee 2 must learn language 2.
import java.io.BufferedReader;
import java.io.FileReader;
import java.io.IOException; 
import java.io.InputStreamReader; 
import java.util.Scanner; 
import java.util.StringTokenizer; 
import java.util.*;
import java.io.*;
public class codeforces 
{ 
   static class Student{
        double e;
       String s;
        Student(String s,double e){
            this.s=s;
            this.e=e;
        }
    }
    static int sieveOfEratosthenes(int n) 
    { 
        // Create a boolean array "prime[0..n]" and initialize 
        // all entries it as true. A value in prime[i] will 
        // finally be false if i is Not a prime, else true. 
        int pos=0;
        boolean prime[] = new boolean[n+1]; 
        for(int i=0;i<=n;i++) 
            prime[i] = true; 
          
        for(int p = 2; p*p <=n; p++) 
        { 
            // If prime[p] is not changed, then it is a prime 
            if(prime[p] == true) 
            { 
                // Update all multiples of p 
                for(int i = p*p; i <= n; i += p) 
                    prime[i] = false; 
            } 
        } 
          
        // Print all prime numbers 
        for(int i = 2; i <= n; i++) 
        { 
            if(prime[i] == true) 
                ++pos;
        }
        return pos;
    }
static    class Sortbyroll implements Comparator<Student> 
{ 
    // Used for sorting in ascending order of 
    // roll number 
    public int compare(Student c, Student b) 
    { 
        String s1,s2;
        s1=c.s.toLowerCase();
        s2=b.s.toLowerCase();
    if(c.e>b.e)
    return -1;
    if(c.e<b.e)
    return 1;
    return s1.compareTo(s2);
    } 
} 
    static class FastReader 
    { 
        BufferedReader br; 
        StringTokenizer st; 
  
        public FastReader() 
        { 
            br = new BufferedReader(new
                     InputStreamReader(System.in)); 
        } 
  
        String next() 
        { 
            while (st == null || !st.hasMoreElements()) 
            { 
                try
                { 
                    st = new StringTokenizer(br.readLine()); 
                } 
                catch (IOException  e) 
                { 
                    e.printStackTrace(); 
                } 
            } 
            return st.nextToken(); 
        } 
  
        int nextInt() 
        { 
            return Integer.parseInt(next()); 
        } 
  
        long nextLong() 
        { 
            return Long.parseLong(next()); 
        } 
  
        double nextDouble() 
        { 
            return Double.parseDouble(next()); 
        } 
  
        String nextLine() 
        { 
            String str = ""; 
            try
            { 
                str = br.readLine(); 
            } 
            catch (IOException e) 
            { 
                e.printStackTrace(); 
            } 
            return str; 
        } 
    }
    static class Edge{
        int a,b;
        Edge(int a,int b){
            this.a=a;
            this.b=b;
        }
    }
    public static void main(String[] args){
        //long sum=0;
        int t,n,d,p,r,j,c_0=-1,c_1=-1,si=0,i,ans2=0,e,mid,min,h,c_2,max,m,pos,temp1,temp2,x,y,c=0,temp,q,k;
        //long ans=0,ans1=0,c_a=0,c_b=0,m,pos,a,b,x,y,z,temp,c;
        double d1,d2,total=0;
        long a1=0,a2=0,ans=1,sum=0,b;
        String s1;
        FastReader sc=new FastReader();
        n=sc.nextInt();
        m=sc.nextInt();
        int parent[]=new int[n+1];
        int a[][]=new int[n][m+1];
        for(i=0;i<n;i++){
            x=sc.nextInt();
            for(j=0;j<x;j++){
                a[i][0]+=1;
                a[i][sc.nextInt()]=1;
            }
        }
        for(i=1;i<=n;i++)
            parent[i]=i;
        for(i=0;i<n;i++){
            x=find(i+1,parent);
            for(j=0;j<n;j++){
                y=find(j+1,parent);
                for(k=1;k<=m;k++){
                    if(a[i][k]!=0&&a[j][k]!=0){
                        if(x<y)
                            parent[y]=x;
                        else
                            parent[x]=y;
                        break;
                    }
                }
            }
        }
        ans=0;
        for(i=1;i<=n;i++){
            //System.out.println(parent[i]);
            if(parent[i]==i&&a[i-1][0]!=0)
                ++ans;
        }
        if(ans!=0)
        ans-=1;
        for(i=0;i<n;i++){
            if(a[i][0]==0)
                ++ans;
        }
        System.out.println(ans);
    }
 static long power(long x, long y, long p) 
{ 
    long res = 1;      // Initialize result 
  
    x = x % p;  // Update x if it is more than or  
                // equal to p 
  
    while (y > 0) 
    { 
        // If y is odd, multiply x with result 
        if ((y & (long)1)%2!=0) 
            res = (res*x) % p; 
  
        // y must be even now 
        y = y>>1; // y = y/2 
        x = (x*x) % p;   
    } 
    return res%p; 
} 
   static int find(int x,int parent[]) 
    { 
        // Finds the representative of the set 
        // that x is an element of 
        while(parent[x]!=x) 
        { 
            // if x is not the parent of itself 
            // Then x is not the representative of 
            // his set, 
            x=parent[x];
  
            // so we recursively call Find on its parent 
            // and move i's node directly under the 
            // representative of this set 
        } 
  
        return x; 
    }
    static void union(int x, int y,int rank[],int parent[]) 
    { 
        // Find representatives of two sets 
        int xRoot = find(x,parent), yRoot = find(y,parent); 
  
        // Elements are in the same set, no need 
        // to unite anything. 
        if (xRoot == yRoot) 
            return; 
  
         // If x's rank is less than y's rank 
        if (rank[xRoot] < rank[yRoot]) 
  
            // Then move x under y  so that depth 
            // of tree remains less 
            parent[xRoot] = yRoot; 
  
        // Else if y's rank is less than x's rank 
        else if (rank[yRoot] < rank[xRoot]) 
  
            // Then move y under x so that depth of 
            // tree remains less 
            parent[yRoot] = xRoot; 
  
        else // if ranks are the same 
        { 
            // Then move y under x (doesn't matter 
            // which one goes where) 
            parent[yRoot] = xRoot; 
  
            // And increment the the result tree's 
            // rank by 1 
            rank[xRoot] = rank[xRoot] + 1; 
        } 
    }  
    static long gcd(long a, long b) 
    { 
        if (a == 0) 
            return b; 
        return gcd(b % a, a); 
    }
}