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 aifc
  import os, sys
  import math
  from io import BytesIO, IOBase
  from collections import Counter
  
  # Fast IO Region
  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")
  
  
  def sieve(n):
      p = [1 for i in range(n + 1)]
      z = 2
      while z * z <= n:
          if p[z] == 1:
              for i in range(z * z, n + 1, z):
                  p[i] = 0
          z += 1
      primes = [2]
      for i in range(3, n + 1, 2):
          if p[i]:
              primes.append(i)
      return primes
  def bins(arr, n, k):
      l = 0
      r = n - 1
      leftGreater = n
      while (l <= r):
          m = int(l + (r - l) / 2)
          if (arr[m] > k):
              leftGreater = m
              r = m - 1
          else:
              l = m + 1
      return leftGreater
  def maxPrimeFactors(n):
      maxPrime = -1
      while n % 2 == 0:
          maxPrime = 2
          n >>= 1
      for i in range(3, int(math.sqrt(n)) + 1, 2):
          while n % i == 0:
              maxPrime = i
              n = n / i
      if n > 2:
          maxPrime = n
      return int(maxPrime)
  def decimalToBinary(N):
      # To store the binary number
      B_Number = 0
      cnt = 0
      while (N != 0):
          rem = N % 2
          c = pow(10, cnt)
          B_Number += rem * c
          N //= 2
          # Count used to store exponent value
          cnt += 1
      return B_Number
  def binaryToDecimal(n):
      num = n
      dec_value = 0
      # Initializing base
      # value to 1, i.e 2 ^ 0
      base1 = 1
      len1 = len(num)
      for i in range(len1 - 1, -1, -1):
          if (num[i] == '1'):
              dec_value += base1
          base1 = base1 * 2
      return dec_value
  def binarySearch(arr, l, r, x):
      while l <= r:
          mid = l + (r - l) // 2;
          # Check if x is present at mid
          if arr[mid] == x:
              return mid
          # If x is greater, ignore left half
          elif arr[mid] < x:
              l = mid + 1
          # If x is smaller, ignore right half
          else:
              r = mid - 1
      # If we reach here, then the element
      # was not present
      return -1
  def merge(arr, l, m, r):
      n1 = m - l + 1
      n2 = r - m
      # create temp arrays
      L = [0] * (n1)
      R = [0] * (n2)
      # Copy data to temp arrays L[] and R[]
      for i in range(0, n1):
          L[i] = arr[l + i]
  
      for j in range(0, n2):
          R[j] = arr[m + 1 + j]
  
      # Merge the temp arrays back into arr[l..r]
      i = 0  # Initial index of first subarray
      j = 0  # Initial index of second subarray
      k = l  # Initial index of merged subarray
  
      while i < n1 and j < n2:
          if L[i] <= R[j]:
              arr[k] = L[i]
              i += 1
          else:
              arr[k] = R[j]
              j += 1
          k += 1
  
      # Copy the remaining elements of L[], if there
      # are any
      while i < n1:
          arr[k] = L[i]
          i += 1
          k += 1
  
      # Copy the remaining elements of R[], if there
      # are any
      while j < n2:
          arr[k] = R[j]
          j += 1
          k += 1
  
  
  # l is for left index and r is right index of the
  # sub-array of arr to be sorted
  def mergeSort(arr, l, r):
      if l < r:
          # Same as (l+r)//2, but avoids overflow for
          # large l and h
          m = (l + (r - 1)) // 2
  
          # Sort first and second halves
          mergeSort(arr, l, m)
          mergeSort(arr, m + 1, r)
          merge(arr, l, m, r)
  def gcd(a, b):
      if a == 0:
          return b
      elif b==0:
          return a
      return gcd(b % a,a)
  
  
  for _ in range(int(input())):
      x,y=map(int,input().split())
      if x==y==0:
          print(0)
      elif x==y:
          print(1)
      else:
          if abs(x-y)==1:
              print(-1)
          elif x==0 or y==0:
              if (x+y)%2!=0:
                  print(-1)
              else:
                  print(2)
          else:
              if(x+y)%2==0:
                  print(2)
              else:
                  print(-1)
  
  
  