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 A {
    static PrintWriter out = new PrintWriter((System.out));
    static Reader sc = new Reader();
  
    public static void main(String args[]) throws IOException {
  
      int t = sc.nextInt();
      while (t-- > 0) {
        solve();
      }
      out.close();
    }
  
    public static void solve() {
      int c = sc.nextInt();
      int d = sc.nextInt();
      int a = 0;
      int b = 0;
      if (c == 0 && d == 0) {
        out.println(0);
      } else if (c == d) {
        out.println(1);
      } else if ((c % 2 != 0 && d % 2 != 0) || (c % 2 == 0 && d % 2 == 0)) {
        out.println(2);
      } else {
        out.println(-1);
      }
    }
  
    static class Reader {
      BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
      StringTokenizer st = new StringTokenizer("");
  
      public String next() {
        while (!st.hasMoreTokens()) {
          try {
            st = new StringTokenizer(br.readLine());
          } catch (Exception e) {
            e.printStackTrace();
          }
        }
        return st.nextToken();
      }
  
      public int nextInt() {
        return Integer.parseInt(next());
      }
  
      public long nextLong() {
        return Long.parseLong(next());
      }
  
      public double nextDouble() {
        return Double.parseDouble(next());
      }
  
      public String nextLine() {
        try {
          return br.readLine();
        } catch (Exception e) {
          e.printStackTrace();
        }
        return null;
      }
  
      public boolean hasNext() {
        String next = null;
        try {
          next = br.readLine();
        } catch (Exception e) {
        }
        if (next == null) {
          return false;
        }
        st = new StringTokenizer(next);
        return true;
      }
    }
  }