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 java.io.*;
  import java.util.*;
  
  public class DeltixSU21A {
  	public static void solve(IO io) {
  
  		int c = io.getInt();
  		int d = io.getInt();
  		
  		if (c == 0 && d == 0) {
  			io.println(0);
  		} else if (c == d) {
  			io.println(1);
  		} else if (c % 2 == d % 2) {
  			io.println(2);
  		} else {
  			io.println(-1);
  		}
  		
  	}
  
  	public static void main(String[] args) {
  		IO io = new IO();
  		int t = io.getInt();
  		for (int testIndex = 0; testIndex < t; testIndex++) {
  			solve(io);
  		}
  		io.close();
  	}
  
  	static class IO extends PrintWriter {
  
  		public IO() {
  			super(new BufferedOutputStream(System.out));
  			r = new BufferedReader(new InputStreamReader(System.in));
  		}
  
  		public IO(InputStream i) {
  			super(new BufferedOutputStream(System.out));
  			r = new BufferedReader(new InputStreamReader(i));
  		}
  
  		public IO(InputStream i, OutputStream o) {
  			super(new BufferedOutputStream(o));
  			r = new BufferedReader(new InputStreamReader(i));
  		}
  
  		public boolean hasMoreTokens() {
  			return peekToken() != null;
  		}
  
  		public int getInt() {
  			return Integer.parseInt(nextToken());
  		}
  
  		public double getDouble() {
  			return Double.parseDouble(nextToken());
  		}
  
  		public long getLong() {
  			return Long.parseLong(nextToken());
  		}
  
  		public String getWord() {
  			return nextToken();
  		}
  
  		private BufferedReader r;
  		private String line;
  		private StringTokenizer st;
  		private String token;
  
  		private String peekToken() {
  			if (token == null)
  				try {
  					while (st == null || !st.hasMoreTokens()) {
  						line = r.readLine();
  						if (line == null)
  							return null;
  						st = new StringTokenizer(line);
  					}
  					token = st.nextToken();
  				} catch (IOException e) {
  				}
  			return token;
  		}
  
  		private String nextToken() {
  			String ans = peekToken();
  			token = null;
  			return ans;
  		}
  	}
  }
  