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;
  template <typename Arg1>
  void __f(const char* name, Arg1&& arg1) {
    cerr << name << " : " << arg1 << '\n';
  }
  template <typename Arg1, typename... Args>
  void __f(const char* names, Arg1&& arg1, Args&&... args) {
    const char* comma = strchr(names + 1, ',');
    cerr.write(names, comma - names) << " : " << arg1 << " | ";
    __f(comma + 1, args...);
  }
  const long long mxN = 2e5 + 100;
  string s;
  long long a[mxN];
  vector<long long> v[mxN];
  long long n, m, k, q;
  long long dp[mxN];
  long long ans;
  long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
  void test_case() {
    long long a, b;
    cin >> a >> b;
    if (a < b) {
      swap(a, b);
    }
    if ((a + b) % 2 == 1) {
      cout << -1 << '\n';
    } else if (a == b) {
      if (a != 0) {
        a = 1;
      }
      cout << a << '\n';
    } else {
      cout << 2 << '\n';
    }
  }
  int32_t main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    long long t;
    t = 1;
    cin >> t;
    for (long long i = 1; i <= t; i++) {
      test_case();
    }
    return 0;
  }
  