description: When you play the game of thrones, you win, or you die. There is no middle ground.
  
  Cersei Lannister, A Game of Thrones by George R. R. Martin
  
  There are n nobles, numbered from 1 to n. Noble i has a power of i. There are also m "friendships". A friendship between nobles a and b is always mutual.
  
  A noble is defined to be vulnerable if both of the following conditions are satisfied: 
  
    * the noble has at least one friend, and 
    * all of that noble's friends have a higher power. 
  
  
  
  You will have to process the following three types of queries. 
  
    1. Add a friendship between nobles u and v. 
    2. Remove a friendship between nobles u and v. 
    3. Calculate the answer to the following process. 
  
  
  
  The process: all vulnerable nobles are simultaneously killed, and all their friendships end. Then, it is possible that new nobles become vulnerable. The process repeats itself until no nobles are vulnerable. It can be proven that the process will end in finite time. After the process is complete, you need to calculate the number of remaining nobles.
  
  Note that the results of the process are not carried over between queries, that is, every process starts with all nobles being alive!
  
  Input
  
  The first line contains the integers n and m (1 ≤ n ≤ 2⋅ 10^5, 0 ≤ m ≤ 2⋅ 10^5) — the number of nobles and number of original friendships respectively.
  
  The next m lines each contain the integers u and v (1 ≤ u,v ≤ n, u ≠ v), describing a friendship. No friendship is listed twice.
  
  The next line contains the integer q (1 ≤ q ≤ 2⋅ {10}^{5}) — the number of queries. 
  
  The next q lines contain the queries themselves, each query has one of the following three formats. 
  
    * 1 u v (1 ≤ u,v ≤ n, u ≠ v) — add a friendship between u and v. It is guaranteed that u and v are not friends at this moment. 
    * 2 u v (1 ≤ u,v ≤ n, u ≠ v) — remove a friendship between u and v. It is guaranteed that u and v are friends at this moment. 
    * 3 — print the answer to the process described in the statement. 
  
  Output
  
  For each type 3 query print one integer to a new line. It is guaranteed that there will be at least one type 3 query.
  
  Examples
  
  Input
  
  
  4 3
  2 1
  1 3
  3 4
  4
  3
  1 2 3
  2 3 1
  3
  
  
  Output
  
  
  2
  1
  
  
  Input
  
  
  4 3
  2 3
  3 4
  4 1
  1
  3
  
  
  Output
  
  
  1
  
  Note
  
  Consider the first example. In the first type 3 query, we have the diagram below.
  
  In the first round of the process, noble 1 is weaker than all of his friends (2 and 3), and is thus killed. No other noble is vulnerable in round 1. In round 2, noble 3 is weaker than his only friend, noble 4, and is therefore killed. At this point, the process ends, and the answer is 2.
  
  <image>
  
  In the second type 3 query, the only surviving noble is 4.
  
  The second example consists of only one type 3 query. In the first round, two nobles are killed, and in the second round, one noble is killed. The final answer is 1, since only one noble survives.
  
  <image>
solution: import java.io.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import static java.lang.Math.*;
  import static java.lang.System.out;
  import java.util.*;
  import java.io.File;
  import java.io.PrintStream;
  import java.io.PrintWriter;
  import java.math.BigInteger;
  public class Main {
   
  	
  	/* 10^(7) = 1s.
  	 * ceilVal = (a+b-1) / b */
  	
  	static final int mod = 1000000007;
  	static final long temp = 998244353;
  	static final long MOD = 1000000007;
  	static final long M = (long)1e9+7;
   
  	static class Pair implements Comparable<Pair> 
  	{
  		int first, second;
  		
  		public Pair(int first, int second) 
  		{
  			this.first = first;
  			this.second = second;
  		}
  		
  		public int compareTo(Pair ob) 
  		{
  			return (int)(first - ob.first);
  		}
  	}
  	
  	
  	static class Tuple implements Comparable<Tuple>
  	{
  		int first, second,third;
  		public Tuple(int first, int second, int third)
  		{
  			this.first = first;
  			this.second = second;
  			this.third = third;
  		}
  		public int compareTo(Tuple o)
  		{
  			return (int)(o.third - this.third);
  		}
  	}
  	
  	public static class DSU
  	{
  		int[] parent;
  		int[] rank; //Size of the trees is used as the rank
  		public DSU(int n)
  		{
  			parent = new int[n];
  			rank = new int[n];
  			Arrays.fill(parent, -1);
  			Arrays.fill(rank, 1);
  		}
  		
  		public int find(int i) //finding through path compression
  		{
  			return parent[i] < 0 ? i : (parent[i] = find(parent[i]));
  		}
  		
  		public boolean union(int a, int b) //Union Find by Rank 
  		{
  			a = find(a);
  			b = find(b);
  			
  			if(a == b) return false; //if they are already connected we exit by returning false.
  			
  			// if a's parent is less than b's parent
  			if(rank[a] < rank[b])
  			{
  				//then move a under b
  				parent[a] = b;
  			}
  			//else if rank of j's parent is less than i's parent
  			else if(rank[a] > rank[b])
  			{
  				//then move b under a
  				parent[b] = a;
  			}
  			//if both have the same rank.
  			else
  			{
  				//move a under b (it doesnt matter if its the other way around.
  				parent[b] = a;
  				rank[a] = 1 + rank[a];
  			}
  			return true;
  		}	
  	}
  
  	static class Reader 
  	{
  		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
  		StringTokenizer st=new StringTokenizer("");
  		String next() {
  			while (!st.hasMoreTokens())
  				try {
  					st=new StringTokenizer(br.readLine());
  				} catch (IOException e) {
  					e.printStackTrace();
  				}
  			return st.nextToken();
  		}
  		
  		int nextInt() {
  			return Integer.parseInt(next());
  		}
  		
  		int[] readArray(int n) throws IOException {
  			int[] a=new int[n];
  			for (int i=0; i<n; i++) a[i]=nextInt();
  			return a;
  		}
  		
  		long[] longReadArray(int n) throws IOException {
  			long[] a=new long[n];
  			for (int i=0; i<n; i++) a[i]=nextLong();
  			return a;
  		}
  		
  		long nextLong() {
  			return Long.parseLong(next());
  		}
  		double nextDouble() {
  			return Double.parseDouble(next());
  		}
  	}
  	
  	public static int gcd(int a, int b)
  	{
  		if(b == 0)
  		 return a;
  		
  		else
  		return gcd(b,a%b);
  	} 
   
  		public static long lcm(long a, long b)
  	    {
  	        return (a / LongGCD(a, b)) * b;
  	    }
  	
  	public static long LongGCD(long a, long b)
  	{
  		if(b == 0)
  			 return a;
  			
  			else
  			return LongGCD(b,a%b);
  	}
  	
  		public static long LongLCM(long a, long b)
  	    {
  	        return (a / LongGCD(a, b)) * b;
  	    }
  	
  		//Count the number of coprime's upto N
  		public static long phi(long n)  //euler totient/phi function
  	    { 
  	       long ans = n;
  //	       for(long i = 2;i*i<=n;i++)
  //	       {
  //	    	   if(n%i == 0)
  //	    	   {
  //	    		   while(n%i == 0) n/=i;
  //	    		   ans -= (ans/i);
  //	    	   }
  //	       }
  //	       
  //	       if(n > 1)
  //	       {
  //	    	   ans -= (ans/n);
  //	       }
  	       
  	       for(long i = 2;i<=n;i++)
  	       {
  	    	   if(isPrime(i))
  	    	   {
  	    		   ans -= (ans/i);
  	    	   }
  	       }
  	       
  	       return ans;
  	       
  	    }	
  	
  	public static long fastPow(long x, long n) 
  	{
  		if(n == 0)
  			return 1;
  		else if(n%2 == 0)
  			return fastPow(x*x,n/2);
  		else
  			return x*fastPow(x*x,(n-1)/2);
  	}
  	
  	   public static long modPow(long x, long y, long p)
  	    {
  	        long res = 1;
  	        x = x % p;
  	 
  	        while (y > 0) {
  	            if (y % 2 == 1)
  	                res = (res * x) % p;
  	 
  	            y = y >> 1; 
  	            x = (x * x) % p;
  	        }
  	        return res;
  	    }
  	   
  	   static long modInverse(long n, long p)
  	    {
  	        return modPow(n, p - 2, p);
  	    }
  	 
  	    // Returns nCr % p using Fermat's little theorem.
  	    
  	    public static long nCrModP(long n, long r,long p)
  	    {
  	    	if (n<r) 
  	              return 0;
  	          
  	        if (r == 0)
  	            return 1;
  	 
  	        long[] fac = new long[(int)(n) + 1];
  	        fac[0] = 1;
  	 
  	        for (int i = 1; i <= n; i++)
  	            fac[i] = fac[i - 1] * i % p;
  	 
  	        return (fac[(int)(n)] * modInverse(fac[(int)(r)], p)
  	                % p * modInverse(fac[(int)(n - r)], p)
  	                % p)
  	            % p;
  	    }
  	    
  	    public static long fact(long n) 
  	    {
  	    	  long[] fac = new long[(int)(n) + 1];
  		        fac[0] = 1;
  		 
  		        for (long i = 1; i <= n; i++)
  		            fac[(int)(i)] = fac[(int)(i - 1)] * i;
  		        
  		        return fac[(int)(n)];
  		}
  	    
  	    public static long nCr(long n, long k) 
  	    {
  			long ans = 1;
  			for(long i = 0;i<k;i++)
  			{
  				ans *= (n-i);
  				ans /= (i+1);
  			}
  			return ans;
  		}
   
  		//Modular Operations for Addition and Multiplication. 
  		   public static long perfomMod(long x)
  		   {
  		        return ((x%M + M)%M);
  		    }
  		   public static long addMod(long a, long b)
  		   {
  		        return perfomMod(perfomMod(a)+perfomMod(b));
  		    }
  		  
  		   public static long subMod(long a, long b)
  		   {
  		        return perfomMod(perfomMod(a)-perfomMod(b));
  		   }
  		   
  		   public static long mulMod(long  a, long b)
  		   {
  		        return perfomMod(perfomMod(a)*perfomMod(b));
  		   }
  		   
  		   public static boolean isPrime(long n) 
  			{
  				if(n == 1)
  				{
  					return false;
  				}
  				for(int i = 2;i*i<=n;i++)
  				{
  					if(n%i == 0)
  					{
  						return false;
  					}
  				}
  				return true;
  			}
  			
  			public static List<Long> SieveList(int n)
  			{
  				boolean prime[] = new boolean[(int)(n+1)];
  				Arrays.fill(prime, true);
  				
  				List<Long> l = new ArrayList<>();
  				for (long p = 2; p*p<=n; p++) 
  				{ 
  					if (prime[(int)(p)] == true) 
  					{ 	
  					    for(long i = p*p; i<=n; i += p) 
  						{
  						    prime[(int)(i)] = false; 
  						}	
  					} 
  				} 
  		 
  				for (long p = 2; p<=n; p++) 
  				{
  				    if (prime[(int)(p)] == true)
  				    {
  				       l.add(p); 
  				    }
  				}
  				
  				return l;
  			}
  
  			public static long log2(long n)
  		    {
  				 long ans = (long)(log(n)/log(2));
  				 return ans;
  			}
  			   
  			public static boolean isPow2(long n)
  			{
  				 return (n  != 0 && ((n & (n-1))) == 0);
  			}	
  		   
  		   public static boolean isSq(int x)  
  		   {
  			    long s = (long)Math.round(Math.sqrt(x));
  			    return s*s==x;
  		   }
  		
  			   /*
  			    * 
  			    * >=         <=
  			   0 1 2 3 4 5 6 7  
  			   5 5 5 6 6 6 7 7
  			   lower_bound for 6 at index 3 (>=)
  			   upper_bound for 6 at index 6(To get six reduce by one) (<=)
  			   */
  
  		   public static int LowerBound(long a[], long x) 
  		   {
  			      int l=-1,r=a.length;
  				  while(l+1<r) 
  				  {
  				    int m=(l+r)>>>1;
  				    if(a[m]>=x) r=m;
  				    else l=m;
  				  }
  				  return r;
  			}
  		   
  			public static int UpperBound(long a[], long x) 
  			{
  				int l=-1, r=a.length;
  			    while(l+1<r) 
  			    {
  			       int m=(l+r)>>>1;
  			       if(a[m]<=x) l=m;
  			       else r=m;
  			    }
  			    return l+1;
  			}
  			
  			public static void Sort(long[] a) 
  			{
  				List<Long> l = new ArrayList<>();
  				for (long i : a) l.add(i);
  				Collections.sort(l);
  //				Collections.reverse(l);  //Use to Sort decreasingly 
  				for (int i=0; i<a.length; i++) a[i]=l.get(i);
  			}
  			
  			public static void ssort(char[] a)
  			{
  				List<Character> l = new ArrayList<>();
  				for (char i : a) l.add(i);
  				Collections.sort(l);
  				for (int i=0; i<a.length; i++) a[i]=l.get(i);
  			}
  			
  		
  	public static int[] map;
  	public static int n,m;
  	public static long c;
  	
  	public static void combine(int a, int b)
  	{
  		int x = min(a,b);
  		if(map[x] == 0) c--;
  		map[x]++;
  	}
  	
  	public static void cut(int a, int b)
  	{
  		int x = min(a,b);
  		if(map[x] == 1) c++;
  		map[x]--;
  	}
  			
  	public static void main(String[] args) throws Exception 
  	{
  		Reader sc = new Reader();
  		PrintWriter fout = new PrintWriter(System.out);
  		
  		n = sc.nextInt();
  		m = sc.nextInt();
  		
  		map = new int[n];
  		c = n;
  		
  		for(int i = 0;i<m;i++)
  		{
  			int x = sc.nextInt()-1, y = sc.nextInt()-1;
  			combine(x,y);
  		}
  		
  		int q  = sc.nextInt();
  		for(int i = 0;i<q;i++)
  		{
  			int query = sc.nextInt();
  			if(query == 1)
  			{
  				int x = sc.nextInt()-1, y = sc.nextInt()-1;
  				combine(x,y);
  			}
  			else if(query == 2)
  			{
  				int x = sc.nextInt()-1, y = sc.nextInt()-1;
  				cut(x,y);
  			}
  			else
  			{
  				fout.println(c);
  			}
  		}
  
  		fout.close();
  	}
  }