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;
  const char nl = '\n';
  const int MAX_N = 100011;
  const long long INF = (1 << 29) + 123;
  const long long MOD = 1000000007;
  const long double PI = 4 * atan((long double)1);
  template <typename T>
  bool ckmin(T& a, const T& b) {
    return a > b ? a = b, 1 : 0;
  }
  template <typename T>
  bool ckmax(T& a, const T& b) {
    return b > a ? a = b, 1 : 0;
  }
  void dbg_out() { cerr << endl; }
  template <typename Head, typename... Tail>
  void dbg_out(Head H, Tail... T) {
    cerr << H << " ";
    dbg_out(T...);
  }
  template <typename T>
  class SafeVector : public vector<T> {
   public:
    using vector<T>::vector;
    typename vector<T>::reference operator[](size_t n) {
      return vector<T>::at(n);
    }
    typename vector<T>::const_reference operator[](size_t n) const {
      return vector<T>::at(n);
    }
  };
  mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
  const int MX = 1 << 20;
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int t;
    cin >> t;
    while (t--) {
      int c, d;
      cin >> c >> d;
      if ((c - d) % 2)
        cout << -1 << nl;
      else if (c == d && c == 0)
        cout << 0 << nl;
      else if (c == d)
        cout << 1 << nl;
      else
        cout << 2 << nl;
    }
    return 0;
  }
  