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: # DEFINING SOME GOOD STUFF
  import heapq
  import sys
  from math import *
  import threading
  from heapq import *
  from itertools import count
  from pprint import pprint
  from collections import defaultdict
  from heapq import heapify, heappop, heappush
  
  
  # threading.stack_size(10**8)
  # sys.setrecursionlimit(300000)
  '''
  -> if you are increasing recursionlimit then remember submitting using python3 rather pypy3
  -> sometimes increasing stack size don't work locally but it will work on CF
  '''
  
  mod = 10 ** 9+7
  inf = 10 ** 15
  decision = ['NO', 'YES']
  yes = 'YES'
  no = 'NO'
  
  # ------------------------------FASTIO----------------------------
  import os
  
  from io import BytesIO, IOBase
  
  BUFSIZE = 8192
  
  
  class FastIO(IOBase):
      newlines = 0
  
      def __init__(self, file):
          self._fd = file.fileno()
          self.buffer = BytesIO()
          self.writable = "x" in file.mode or "r" not in file.mode
          self.write = self.buffer.write if self.writable else None
  
      def read(self):
          while True:
              b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
              if not b:
                  break
              ptr = self.buffer.tell()
              self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
          self.newlines = 0
          return self.buffer.read()
  
      def readline(self):
          while self.newlines == 0:
              b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
              self.newlines = b.count(b"\n")+(not b)
              ptr = self.buffer.tell()
              self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
          self.newlines -= 1
          return self.buffer.readline()
  
      def flush(self):
          if self.writable:
              os.write(self._fd, self.buffer.getvalue())
              self.buffer.truncate(0), self.buffer.seek(0)
  
  
  class IOWrapper(IOBase):
      def __init__(self, file):
          self.buffer = FastIO(file)
          self.flush = self.buffer.flush
          self.writable = self.buffer.writable
          self.write = lambda s: self.buffer.write(s.encode("ascii"))
          self.read = lambda: self.buffer.read().decode("ascii")
          self.readline = lambda: self.buffer.readline().decode("ascii")
  
  
  sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
  input = lambda: sys.stdin.readline().rstrip("\r\n")
  
  
  # ________________________FAST FACTORIAL______________________________#
  class Factorial:
      def __init__(self, MOD):
          self.MOD = MOD
          self.factorials = [1, 1]
          self.invModulos = [0, 1]
          self.invFactorial_ = [1, 1]
  
      def calc(self, n):
          if n <= -1:
              print("Invalid argument to calculate n!")
              print("n must be non-negative value. But the argument was "+str(n))
              exit()
          if n < len(self.factorials):
              return self.factorials[n]
          nextArr = [0] * (n+1-len(self.factorials))
          initialI = len(self.factorials)
          prev = self.factorials[-1]
          m = self.MOD
          for i in range(initialI, n+1):
              prev = nextArr[i-initialI] = prev * i % m
          self.factorials += nextArr
          return self.factorials[n]
  
      def inv(self, n):
          if n <= -1:
              print("Invalid argument to calculate n^(-1)")
              print("n must be non-negative value. But the argument was "+str(n))
              exit()
          p = self.MOD
          pi = n % p
          if pi < len(self.invModulos):
              return self.invModulos[pi]
          nextArr = [0] * (n+1-len(self.invModulos))
          initialI = len(self.invModulos)
          for i in range(initialI, min(p, n+1)):
              next = -self.invModulos[p % i] * (p // i) % p
              self.invModulos.append(next)
          return self.invModulos[pi]
  
      def invFactorial(self, n):
          if n <= -1:
              print("Invalid argument to calculate (n^(-1))!")
              print("n must be non-negative value. But the argument was "+str(n))
              exit()
          if n < len(self.invFactorial_):
              return self.invFactorial_[n]
          self.inv(n)  # To make sure already calculated n^-1
          nextArr = [0] * (n+1-len(self.invFactorial_))
          initialI = len(self.invFactorial_)
          prev = self.invFactorial_[-1]
          p = self.MOD
          for i in range(initialI, n+1):
              prev = nextArr[i-initialI] = (prev * self.invModulos[i % p]) % p
          self.invFactorial_ += nextArr
          return self.invFactorial_[n]
  
  class Combination:
      def __init__(self, MOD):
          self.MOD = MOD
          self.factorial = Factorial(MOD)
  
      def ncr(self, n, k):
          if k < 0 or n < k:
              return 0
          k = min(k, n-k)
          f = self.factorial
          return f.calc(n) * f.invFactorial(max(n-k, k)) * f.invFactorial(min(k, n-k)) % self.MOD
  
      def npr(self, n, k):
          if k < 0 or n < k:
              return 0
          f = self.factorial
          return (f.calc(n) * f.invFactorial(n-k)) % self.MOD
  
  
  #_______________SEGMENT TREE ( logn range modifications )_____________#
  
  class SegmentTree:
      def __init__(self, data, default = 0, func = lambda a, b: max(a, b)):
          """initialize the segment tree with data"""
          self._default = default
          self._func = func
          self._len = len(data)
          self._size = _size = 1 << (self._len-1).bit_length()
  
          self.data = [default] * (2 * _size)
          self.data[_size:_size+self._len] = data
          for i in reversed(range(_size)):
              self.data[i] = func(self.data[i+i], self.data[i+i+1])
  
      def __delitem__(self, idx):
          self[idx] = self._default
  
      def __getitem__(self, idx):
          return self.data[idx+self._size]
  
      def __setitem__(self, idx, value):
          idx += self._size
          self.data[idx] = value
          idx >>= 1
          while idx:
              self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx+1])
              idx >>= 1
  
      def __len__(self):
          return self._len
  
      def query(self, start, stop):
          if start == stop:
              return self.__getitem__(start)
          stop += 1
          start += self._size
          stop += self._size
  
          res = self._default
          while start < stop:
              if start & 1:
                  res = self._func(res, self.data[start])
                  start += 1
              if stop & 1:
                  stop -= 1
                  res = self._func(res, self.data[stop])
              start >>= 1
              stop >>= 1
          return res
  
      def __repr__(self):
          return "SegmentTree({0})".format(self.data)
  
  
  
  # ____________________MY FAVOURITE FUNCTIONS_______________________#
  
  
  
  def lower_bound(li, num):
      answer = len(li)
      start = 0
      end = len(li)-1
  
      while (start <= end):
          middle = (end+start) // 2
          if li[middle] >= num:
              answer = middle
              end = middle-1
          else:
              start = middle+1
      return answer  # min index where x is not less than num
  
  
  def upper_bound(li, num):
      answer = len(li)
      start = 0
      end = len(li)-1
  
      while (start <= end):
          middle = (end+start) // 2
          if li[middle] <= num:
              start = middle+1
  
          else:
              answer = middle
              end = middle-1
      return answer  # max index where x is greater than num
  
  
  def abs(x):
      return x if x >= 0 else -x
  
  
  def binary_search(li, val):
      # print(lb, ub, li)
      ans = -1
      lb = 0
      ub = len(li)-1
      while (lb <= ub):
          mid = (lb+ub) // 2
          # print('mid is',mid, li[mid])
          if li[mid] > val:
              ub = mid-1
          elif val > li[mid]:
              lb = mid+1
          else:
              ans = mid  # return index
              break
      return ans
  
  
  def kadane(x):  # maximum sum contiguous subarray
      sum_so_far = 0
      current_sum = 0
      for i in x:
          current_sum += i
          if current_sum < 0:
              current_sum = 0
          else:
              sum_so_far = max(sum_so_far, current_sum)
      return sum_so_far
  
  
  def pref(li):
      pref_sum = [0]
      for i in li:
          pref_sum.append(pref_sum[-1]+i)
      return pref_sum
  
  
  def SieveOfEratosthenes(n):
      prime = [{1, i} for i in range(n+1)]
      p = 2
      while (p <= n):
          for i in range(p * 2, n+1, p):
              prime[i].add(p)
          p += 1
      return prime
  
  
  def primefactors(n):
      factors = []
      while (n % 2 == 0):
          factors.append(2)
          n //= 2
      for i in range(3, int(sqrt(n))+1, 2):  # only odd factors left
          while n % i == 0:
              factors.append(i)
              n //= i
      if n > 2:  # incase of prime
          factors.append(n)
      return factors
  
  
  def prod(li):
      ans = 1
      for i in li:
          ans *= i
      return ans
  
  
  def sumk(a, b):
      print('called for', a, b)
      ans = a * (a+1) // 2
      ans -= b * (b+1) // 2
      return ans
  
  def sumi(n):
      ans = 0
      if len(n) > 1:
          for x in n:
              ans += int(x)
          return ans
      else:
          return int(n)
  
  def checkwin(x, a):
      if a[0][0] == a[1][1] == a[2][2] == x:
          return 1
      if a[0][2] == a[1][1] == a[2][0] == x:
          return 1
      if (len(set(a[0])) == 1 and a[0][0] == x) or (len(set(a[1])) == 1 and a[1][0] == x) or (len(set(a[2])) == 1 and a[2][0] == x):
          return 1
      if (len(set(a[0][:])) == 1 and a[0][0] == x) or (len(set(a[1][:])) == 1 and a[0][1] == x) or (len(set(a[2][:])) == 1 and a[0][0] == x):
          return 1
      return 0
  
  # _______________________________________________________________#
  
  
  inf = 10**9 + 7
  
  def main():
      # karmanya = int(input())
      karmanya = 1
      # divisors = SieveOfEratosthenes(200010)
      # print(divisors)
      while karmanya != 0:
          karmanya -= 1
          # n = int(input())
          n,m = map(int, input().split())
          # a = [int(x) for x in list(input())]
          # b = [int(x) for x in list(input())]
          # s = list(input())
          # a = list(map(int, input().split()))
          # b = list(map(int, input().split()))
          # c = list(map(int, input().split()))
          friend = [1]*(n+1)
          ans = n
          for i in range(m):
              k = min(list(map(int, input().split())))
              friend[k] += 1
          for i in range(n+1):
              if friend[i] > 1:
                  ans -= 1
          for q in range(int(input())):
              command = list(map(int, input().split()))
              if len(command) == 3:
                  a,b = command[1], command[2]
                  if command[0] == 1:
                      friend[min(a,b)] += 1
                      if friend[min(a,b)] == 2:
                          ans -= 1
                  else:
                      friend[min(a,b)] -= 1
                      if friend[min(a,b)] == 1:
                          ans += 1
              else:
                  print(ans)
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  main()
  # t = threading.Thread(target=main)
  # t.start()
  # t.join()