A Variety of Operations
  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 $$$k$$$)  add number $$$k$$$ to both $$$a$$$ and $$$b$$$, or  add number $$$k$$$ to $$$a$$$ and subtract $$$k$$$ from $$$b$$$, or  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 $$$d$$$.
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 $$$(3, 5)$$$. 
Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). Description of the test cases follows.The only line of each test case contains two integers $$$c$$$ and $$$d$$$ $$$(0 \le c, d \le 10^9)$$$, which are William's favorite numbers and which he wants $$$a$$$ and $$$b$$$ to be transformed into.
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.
for s in [*open(0)][1:]:
    (c, d) = map(int, s.split())
    print(-(c + d & 1) or (c != d) * 2 or +(c > 0))
import sys
input = sys.stdin.readline

def solve():
    (c, d) = map(int, input().split())
    if c == d:
        return 0 if c == 0 else 1
    elif (c + d) % 2 == 0:
        return 2
    else:
        return -1
for _ in range(int(input())):
    print(solve())
import sys
input = sys.stdin.readline

def main():
    n = int(input())
    for i in range(n):
        (p, q) = map(int, input().split())
        if (p + q) % 2 != 0:
            print(-1)
        elif p == 0 and q == 0:
            print(0)
        elif p == q:
            print(1)
        else:
            print(2)
if __name__ == '__main__':
    main()