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 <class C>
  struct rge {
    C b, e;
  };
  template <class C>
  rge<C> range(C x, C y) {
    return rge<C>{x, y};
  }
  template <class C>
  auto dud(C *x) -> decltype(cerr << *x, 0);
  template <class C>
  char dud(...);
  struct debug {
    ~debug() { cerr << "\n"; }
    template <class C>
    typename enable_if<sizeof dud<C>(0) != 1, debug &>::type operator<<(C i) {
      cerr << boolalpha << i;
      return *this;
    }
    template <class C>
    typename enable_if<sizeof dud<C>(0) == 1, debug &>::type operator<<(C i) {
      return *this << range(begin(i), end(i));
    }
    template <class C1, class C2>
    debug &operator<<(pair<C1, C2> p) {
      return *this << "(" << p.first << ", " << p.second << ")";
    }
    template <class C>
    debug &operator<<(rge<C> r) {
      *this << "[";
      for (auto it = r.b; it != r.e; ++it) *this << ", " + 2 * (it == r.b) << *it;
      return *this << "]";
    }
  };
  long long c, d;
  void test_case() {
    cin >> c >> d;
    if (c == 0 && d == 0) {
      cout << "0\n";
      return;
    }
    if (abs(c - d) & 1) {
      cout << "-1\n";
      return;
    }
    cout << ((c - d == 0) ? 1 : 2) << "\n";
  }
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    int T;
    cin >> T;
    while (T--) {
      test_case();
    }
    return 0;
  }
  