description: <image>
  
  William has two numbers a and b initially both equal to zero. William mastered performing three different operations with them quickly. Before performing each operation some positive integer k is picked, which is then used to perform one of the following operations: (note, that for each operation you can choose a new positive integer k)
  
    1. add number k to both a and b, or 
    2. add number k to a and subtract k from b, or 
    3. add number k to b and subtract k from a. 
  
  
  
  Note that after performing operations, numbers a and b may become negative as well.
  
  William wants to find out the minimal number of operations he would have to perform to make a equal to his favorite number c and b equal to his second favorite number d.
  
  Input
  
  Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4). Description of the test cases follows.
  
  The only line of each test case contains two integers c and d (0 ≤ c, d ≤ 10^9), which are William's favorite numbers and which he wants a and b to be transformed into.
  
  Output
  
  For each test case output a single number, which is the minimal number of operations which William would have to perform to make a equal to c and b equal to d, or -1 if it is impossible to achieve this using the described operations.
  
  Example
  
  Input
  
  
  6
  1 2
  3 5
  5 3
  6 6
  8 0
  0 0
  
  
  Output
  
  
  -1
  2
  2
  1
  2
  0
  
  Note
  
  Let us demonstrate one of the suboptimal ways of getting a pair (3, 5):
  
    * Using an operation of the first type with k=1, the current pair would be equal to (1, 1). 
    * Using an operation of the third type with k=8, the current pair would be equal to (-7, 9). 
    * Using an operation of the second type with k=7, the current pair would be equal to (0, 2). 
    * Using an operation of the first type with k=3, the current pair would be equal to (3, 5). 
solution: import os
  import sys
  from io import BytesIO, IOBase
  from collections import Counter
  import math as mt
  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)
  
  
  # def gcd(a, b):
  #     if a == 0:
  #         return b
  #     return gcd(b % a, a)
  #
  # def lcm(a, b):
  #     return (a * b) / gcd(a, b)
  #
  # mod = int(1e9) + 7
  # def power(k, n):
  #     if n == 0:
  #         return 1
  #     if n % 2:
  #         return (power(k, n - 1) * k) % mod
  #     t = power(k, n // 2)
  #     return (t * t) % mod
  #
  # def totalPrimeFactors(n):
  #     count = 0
  #     if (n % 2) == 0:
  #         count += 1
  #         while (n % 2) == 0:
  #             n //= 2
  #
  #     i = 3
  #     while i * i <= n:
  #         if (n % i) == 0:
  #             count += 1
  #             while (n % i) == 0:
  #                 n //= i
  #         i += 2
  #     if n > 2:
  #         count += 1
  #     return count
  
  # #MAXN = int(1e7 + 1)
  # # spf = [0 for i in range(MAXN)]
  #
  #
  # def sieve():
  #     spf[1] = 1
  #     for i in range(2, MAXN):
  #         spf[i] = i
  #     for i in range(4, MAXN, 2):
  #         spf[i] = 2
  #
  #     for i in range(3, mt.ceil(mt.sqrt(MAXN))):
  #         if (spf[i] == i):
  #             for j in range(i * i, MAXN, i):
  #                 if (spf[j] == j):
  #                     spf[j] = i
  #
  #
  # def getFactorization(x):
  #     ret = 0
  #     while (x != 1):
  #         k = spf[x]
  #         ret += 1
  #         # ret.add(spf[x])
  #         while x % k == 0:
  #             x //= k
  #
  #     return ret
  
  
  # Driver code
  
  # precalculating Smallest Prime Factor
  # sieve()
  
  def main():
      for _ in range(int(input())):
          # n=int(input())
          # a=list(map(int, input().split()))
          a, b = map(int,input().split())
          z=(a+b)//2
          if abs(a-z)!=abs(b-z):
              print(-1)
          else:
              if a==b==0:
                  print(0)
              elif a==z:
                  print(1)
              else:
                  print(2)
  
          #s=input()
  
      return
  
  
  if __name__ == "__main__":
      main()
  