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 static java.lang.Integer.parseInt;
  import static java.lang.Long.parseLong;
  import static java.lang.System.exit;
  
  import java.io.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.io.PrintWriter;
  import java.util.StringTokenizer;
  
  public class A {
  
  	static void solve() throws Exception {
  		int tests = scanInt();
  		for (int test = 0; test < tests; test++) {
  			int c = scanInt(), d = scanInt();
  			out.println(c == 0 && d == 0 ? 0 : c == d ? 1 : ((c ^ d) & 1) == 0 ? 2 : -1);
  		}
  	}
  
  	static int scanInt() throws IOException {
  		return parseInt(scanString());
  	}
  
  	static long scanLong() throws IOException {
  		return parseLong(scanString());
  	}
  
  	static String scanString() throws IOException {
  		while (tok == null || !tok.hasMoreTokens()) {
  			tok = new StringTokenizer(in.readLine());
  		}
  		return tok.nextToken();
  	}
  
  	static BufferedReader in;
  	static PrintWriter out;
  	static StringTokenizer tok;
  
  	public static void main(String[] args) {
  		try {
  			in = new BufferedReader(new InputStreamReader(System.in));
  			out = new PrintWriter(System.out);
  			solve();
  			in.close();
  			out.close();
  		} catch (Throwable e) {
  			e.printStackTrace();
  			exit(1);
  		}
  	}
  }