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.*;
  import java.util.*;
  
  public class start {
  
  	public static void main(String args[]) throws Exception {
  		Scanner sc = new Scanner(System.in);
  		PrintWriter pw = new PrintWriter(System.out);
  
  		int t = 1;
  		while (t-- > 0) {
  			int n = sc.nextInt();
  			int m = sc.nextInt();
  			int a[] = new int[n];
  			HashSet<Integer> hs = new HashSet<Integer>();
  			int r=0;
  			for (int i = 0; i < m; i++) {
  				int u = sc.nextInt() - 1;
  				int v = sc.nextInt() - 1;
  				int x=Math.min(u, v);
  				a[x]++;
  				if(a[x]==1)
  					r++;
  			}
  			int q = sc.nextInt();
  			while (q-- > 0) {
  				int x = sc.nextInt();
  				if (x == 1) {
  					int u = sc.nextInt() - 1;
  					int v = sc.nextInt() - 1;
  					int w=Math.min(u, v);
  					a[w]++;
  					if(a[w]==1)
  						r++;
  				}
  				if (x == 2) {
  					int u = sc.nextInt() - 1;
  					int v = sc.nextInt() - 1;
  					int w=Math.min(u, v);
  					a[w]--;
  					if (a[w] == 0)
  						r--;
  				}
  				if (x == 3)
  					pw.println(n - r);
  			}
  
  		}
  		pw.flush();
  	}
  
  ///////////////////////////////////////////////////////////////////////////////////////////
  
  	static class tri {
  		int x, y, z;
  
  		public tri(int x, int y, int z) {
  			this.x = x;
  			this.y = y;
  			this.z = z;
  		}
  
  	}
  
  	static class pair implements Comparable<pair> {
  		int x, y;
  		boolean w, l;
  		PriorityQueue<Long> pq;
  
  		public pair(boolean a, boolean b) {
  			w = a;
  			l = b;
  		}
  
  		pair(int s, int d) {
  			x = s;
  			y = d;
  		}
  
  		@Override
  		public int compareTo(pair p) {
  			return Long.compare(x, p.x);
  		}
  
  		@Override
  		public String toString() {
  			return x + " " + y;
  
  		}
  
  	}
  
  	static long mod(long ans, int mod) {
  		return (ans % mod + mod) % mod;
  	}
  
  	static long gcd(long a, long b) {
  		if (a == 0)
  			return b;
  		return gcd(b % a, a);
  	}
  
  	static long lcm(long a, long b) {
  		return (a * b) / gcd(a, b);
  	}
  
  	public static int log(int n, int base) {
  		int ans = 0;
  		while (n + 1 > base) {
  			ans++;
  			n /= base;
  		}
  		return ans;
  	}
  
  	static long pow(long b, long e) {
  		long ans = 1;
  		while (e > 0) {
  			if ((e & 1) == 1)
  				ans = ((ans * 1l * b));
  			e >>= 1;
  			{
  
  			}
  			b = ((b * 1l * b));
  		}
  		return ans;
  	}
  
  	static long powmod(long r, long e, int mod) {
  		long ans = 1;
  		r %= mod;
  		while (e > 0) {
  			if ((e & 1) == 1)
  				ans = (int) ((ans * 1l * r) % mod);
  			e >>= 1;
  			r = (int) ((r * 1l * r) % mod);
  		}
  		return ans;
  	}
  
  	static int ceil(int a, int b) {
  		int ans = a / b;
  		return a % b == 0 ? ans : ans + 1;
  	}
  
  	static long ceil(long a, long b) {
  		long ans = a / b;
  		return a % b == 0 ? ans : ans + 1;
  	}
  
  	static HashMap<Integer, Integer> compress(int a[]) {
  		TreeSet<Integer> ts = new TreeSet<>();
  		HashMap<Integer, Integer> hm = new HashMap<>();
  		for (int x : a)
  			ts.add(x);
  		for (int x : ts) {
  			hm.put(x, hm.size() + 1);
  		}
  		return hm;
  	}
  
  	// Returns nCr % p
  	static int C[];
  
  	static int nCrModp(int n, int r, int p) {
  		if (r > n - r)
  			r = n - r;
  		if (C[r] != 0)
  			return C[r];
  		// The array C is going to store last
  		// row of pascal triangle at the end.
  		// And last entry of last row is nCr
  
  		C[0] = 1; // Top row of Pascal Triangle
  
  		// One by constructs remaining rows of Pascal
  		// Triangle from top to bottom
  		for (int i = 1; i <= n; i++) {
  
  			// Fill entries of current row using previous
  			// row values
  			for (int j = Math.min(i, r); j > 0; j--)
  
  				// nCj = (n-1)Cj + (n-1)C(j-1);
  				C[j] = (C[j] + C[j - 1]) % p;
  		}
  		return C[r];
  	}
  
  	static class Scanner {
  		StringTokenizer st;
  		BufferedReader br;
  
  		public Scanner(InputStream s) {
  			br = new BufferedReader(new InputStreamReader(s));
  		}
  
  		public Scanner(String file) throws Exception {
  			br = new BufferedReader(new FileReader(file));
  		}
  
  		public int[] intArr(int n) throws IOException {
  			int a[] = new int[n];
  			for (int i = 0; i < a.length; i++) {
  				a[i] = nextInt();
  			}
  			return a;
  		}
  
  		public long[] longArr(int n) throws IOException {
  			long a[] = new long[n];
  			for (int i = 0; i < a.length; i++) {
  				a[i] = nextLong();
  			}
  			return a;
  		}
  
  		public String next() throws IOException {
  			while (st == null || !st.hasMoreTokens())
  				st = new StringTokenizer(br.readLine());
  			return st.nextToken();
  		}
  
  		public int nextInt() throws IOException {
  			return Integer.parseInt(next());
  		}
  
  		public long nextLong() throws IOException {
  			return Long.parseLong(next());
  		}
  
  		public String nextLine() throws IOException {
  			return br.readLine();
  		}
  
  		public double nextDouble() throws IOException {
  			String x = next();
  			StringBuilder sb = new StringBuilder("0");
  			double res = 0, f = 1;
  			boolean dec = false, neg = false;
  			int start = 0;
  			if (x.charAt(0) == '-') {
  				neg = true;
  				start++;
  			}
  			for (int i = start; i < x.length(); i++)
  				if (x.charAt(i) == '.') {
  					res = Long.parseLong(sb.toString());
  					sb = new StringBuilder("0");
  					dec = true;
  				} else {
  					sb.append(x.charAt(i));
  					if (dec)
  						f *= 10;
  				}
  			res += Long.parseLong(sb.toString()) / f;
  			return res * (neg ? -1 : 1);
  		}
  
  		public boolean ready() throws IOException {
  			return br.ready();
  		}
  
  	}
  
  	public static void shuffle(int[] times2) {
  		int n = times2.length;
  		for (int i = 0; i < n; i++) {
  			int r = i + (int) (Math.random() * (n - i));
  			int tmp = times2[i];
  			times2[i] = times2[r];
  			times2[r] = tmp;
  		}
  	}
  
  }