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>
3 0
2
1 2 1
3
 20 0
1
3
 1000 0
1
3
 20 0
20
3
3
1 17 19
1 19 20
1 4 8
1 9 18
1 1 14
1 16 19
1 10 19
1 1 7
1 6 10
1 18 19
1 9 16
1 15 17
1 11 12
1 10 15
3
1 2 9
1 1 13
1 16 20
 20 53
1 2
1 4
1 5
1 6
1 7
1 8
1 20
2 3
2 6
2 16
2 17
3 6
3 18
3 20
4 6
4 7
4 8
4 9
4 14
4 18
5 11
5 13
5 17
5 18
5 19
6 8
6 13
7 8
7 10
7 15
7 17
7 18
8 15
8 16
9 10
9 15
10 11
10 17
10 20
11 15
11 16
12 14
12 15
12 16
12 19
14 16
14 19
16 17
16 19
16 20
17 19
17 20
18 19
1
3

2
 20
 1000
 20
20
9
 4



import java.util.*;
import java.io.*;

public class C {
	static StringBuilder sb;
	static dsu dsu;
	static long fact[];
	static int mod = (int) (1e9 + 7);
	static int si;
	static int ei;

	static void solve() {
		int n = i();
		int m = i();

		int[] outdegree = new int[n + 1];
		int count = n;
		for (int i = 0; i < m; i++) {
			int u = i();
			int v = i();
			if (u > v) {
				int temp = v;
				v = u;
				u = temp;
			}
			if (outdegree[u] == 0) {
				count--;
			}
			outdegree[u] = outdegree[u] + 1;
		}
		int q = i();
		for (int i = 0; i < q; i++) {
			int num = i();

			if (num == 1) {
				int u = i();
				int v = i();
				if (u > v) {
					int temp = v;
					v = u;
					u = temp;
				}
				if (outdegree[u] == 0) {
					count--;
				}
				outdegree[u] = outdegree[u] + 1;
			} else if (num == 2) {
				int u = i();
				int v = i();
				if (u > v) {
					int temp = v;
					v = u;
					u = temp;
				}
				outdegree[u] = outdegree[u] - 1;
				if (outdegree[u] == 0) {
					count++;
				}
			} else {
				sb.append(count + "\n");
			}
		}
	}

	public static void main(String[] args) {
		sb = new StringBuilder();
		int test = 1;
		while (test-- > 0) {
			solve();
		}
		System.out.println(sb);

	}

	/*
	 * fact=new long[(int)1e6+10]; fact[0]=fact[1]=1; for(int i=2;i<fact.length;i++)
	 * { fact[i]=((long)(i%mod)1L(long)(fact[i-1]%mod))%mod; }
	 */
//**************NCR%P******************	 
	static long ncr(int n, int r) {
		if (r > n)
			return (long) 0;

		long res = fact[n] % mod;
		// System.out.println(res);
		res = ((long) (res % mod) * (long) (p(fact[r], mod - 2) % mod)) % mod;
		res = ((long) (res % mod) * (long) (p(fact[n - r], mod - 2) % mod)) % mod;
		// System.out.println(res);
		return res;

	}

	static long p(long x, long y)// POWER FXN //
	{
		if (y == 0)
			return 1;

		long res = 1;
		while (y > 0) {
			if (y % 2 == 1) {
				res = (res * x) % mod;
				y--;
			}

			x = (x * x) % mod;
			y = y / 2;

		}
		return res;
	}

//**************END******************

	// *************Disjoint set
	// union*********//
	static class dsu {
		int parent[];

		dsu(int n) {
			parent = new int[n];
			for (int i = 0; i < n; i++)
				parent[i] = -1;
		}

		int find(int a) {
			if (parent[a] < 0)
				return a;
			else {
				int x = find(parent[a]);
				parent[a] = x;
				return x;
			}
		}

		void merge(int a, int b) {
			a = find(a);
			b = find(b);
			if (a == b)
				return;
			parent[b] = a;
		}
	}

//**************PRIME FACTORIZE **********************************//
	static TreeMap<Integer, Integer> prime(long n) {
		TreeMap<Integer, Integer> h = new TreeMap<>();
		long num = n;
		for (int i = 2; i <= Math.sqrt(num); i++) {
			if (n % i == 0) {
				int nt = 0;
				while (n % i == 0) {
					n = n / i;
					nt++;
				}
				h.put(i, nt);
			}
		}
		if (n != 1)
			h.put((int) n, 1);
		return h;

	}

//****CLASS PAIR ************************************************
	static class Pair implements Comparable<Pair> {
		int x;
		int y;

		Pair(int x, int y) {
			this.x = x;
			this.y = y;
		}

		public int compareTo(Pair o) {
			return (int) (this.y - o.y);

		}

	}
//****CLASS PAIR **************************************************

	static class InputReader {
		private InputStream stream;
		private byte[] buf = new byte[1024];
		private int curChar;
		private int numChars;
		private SpaceCharFilter filter;

		public InputReader(InputStream stream) {
			this.stream = stream;
		}

		public int read() {
			if (numChars == -1)
				throw new InputMismatchException();
			if (curChar >= numChars) {
				curChar = 0;
				try {
					numChars = stream.read(buf);
				} catch (IOException e) {
					throw new InputMismatchException();
				}
				if (numChars <= 0)
					return -1;
			}
			return buf[curChar++];
		}

		public int Int() {
			int c = read();
			while (isSpaceChar(c))
				c = read();
			int sgn = 1;
			if (c == '-') {
				sgn = -1;
				c = read();
			}
			int res = 0;
			do {
				if (c < '0' || c > '9')
					throw new InputMismatchException();
				res *= 10;
				res += c - '0';
				c = read();
			} while (!isSpaceChar(c));
			return res * sgn;
		}

		public String String() {
			int c = read();
			while (isSpaceChar(c))
				c = read();
			StringBuilder res = new StringBuilder();
			do {
				res.appendCodePoint(c);
				c = read();
			} while (!isSpaceChar(c));
			return res.toString();
		}

		public boolean isSpaceChar(int c) {
			if (filter != null)
				return filter.isSpaceChar(c);
			return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
		}

		public String next() {
			return String();
		}

		public interface SpaceCharFilter {
			public boolean isSpaceChar(int ch);
		}
	}

	static class OutputWriter {
		private final PrintWriter writer;

		public OutputWriter(OutputStream outputStream) {
			writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream)));
		}

		public OutputWriter(Writer writer) {
			this.writer = new PrintWriter(writer);
		}

		public void print(Object... objects) {
			for (int i = 0; i < objects.length; i++) {
				if (i != 0)
					writer.print(' ');
				writer.print(objects[i]);
			}
		}

		public void printLine(Object... objects) {
			print(objects);
			writer.println();
		}

		public void close() {
			writer.close();
		}

		public void flush() {
			writer.flush();
		}
	}

	static InputReader in = new InputReader(System.in);
	static OutputWriter out = new OutputWriter(System.out);

	public static long[] sort(long[] a2) {
		int n = a2.length;
		ArrayList<Long> l = new ArrayList<>();
		for (long i : a2)
			l.add(i);
		Collections.sort(l);
		for (int i = 0; i < l.size(); i++)
			a2[i] = l.get(i);
		return a2;
	}

	public static int[] sort(int[] a2) {
		int n = a2.length;
		ArrayList<Integer> l = new ArrayList<>();
		for (int i : a2)
			l.add(i);
		Collections.sort(l);
		for (int i = 0; i < l.size(); i++)
			a2[i] = l.get(i);
		return a2;
	}

	public static long pow(long x, long y) {
		long res = 1;
		while (y > 0) {
			if (y % 2 != 0) {
				res = (res * x);// % modulus;
				y--;

			}
			x = (x * x);// % modulus;
			y = y / 2;
		}
		return res;
	}

//GCD___+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
	public static long gcd(long x, long y) {
		if (x == 0)
			return y;
		else
			return gcd(y % x, x);
	}
	// ******LOWEST COMMON MULTIPLE
	// *********************************************

	public static long lcm(long x, long y) {
		return (x * (y / gcd(x, y)));
	}

//INPUT PATTERN********************************************************
	public static int i() {
		return in.Int();
	}

	public static long l() {
		String s = in.String();
		return Long.parseLong(s);
	}

	public static String s() {
		return in.String();
	}

	public static int[] readArrayi(int n) {
		int A[] = new int[n];
		for (int i = 0; i < n; i++) {
			A[i] = i();
		}
		return A;
	}

	public static long[] readArray(long n) {
		long A[] = new long[(int) n];
		for (int i = 0; i < n; i++) {
			A[i] = l();
		}
		return A;
	}

}