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: #include <bits/stdc++.h>
  using namespace std;
  inline char gc() {
    static char buf[100000], *p1 = buf, *p2 = buf;
    return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2)
               ? EOF
               : *p1++;
  }
  inline long long read() {
    char c = getchar();
    long long su = 0, f = 1;
    for (; c < '0' || c > '9'; c = getchar())
      if (c == '-') f = -1;
    for (; c >= '0' && c <= '9'; c = getchar()) su = su * 10 + c - '0';
    return su * f;
  }
  inline void write(long long x) {
    if (x < 0) {
      putchar('-');
      write(-x);
      return;
    }
    if (x >= 10) write(x / 10);
    putchar(x % 10 + '0');
  }
  inline void writesp(long long x) { write(x), putchar(' '); }
  inline void writeln(long long x) {
    write(x);
    putchar('\n');
  }
  int T, x, y, n, m;
  signed main() {
    T = read();
    while (T--) {
      x = 0, y = 0, n = read(), m = read();
      if ((x + y) % 2 != (n + m) % 2)
        puts("-1");
      else {
        if (x == n && y == m)
          puts("0");
        else if (n - x == m - y || x + y == n + m)
          puts("1");
        else
          puts("2");
      }
    }
  }
  