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>
  #pragma GCC optimize("Ofast")
  #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,fma")
  #pragma GCC optimize("unroll-loops")
  using namespace std;
  bool per_square(long double a) {
    if (a < 0) return false;
    long long sr = sqrt(a);
    return (sr * sr == a);
  }
  bool is_prime(long long x) {
    for (long long i = 2; i < sqrt(x) + 1; i++)
      if (x % i == 0) return false;
    return true;
  }
  bool sort_by_sec(const pair<long long, long long> &a,
                   const pair<long long, long long> &b) {
    return (a.second < b.second);
  }
  template <typename T>
  bool is_pali(T A) {
    long long n = (long long)((A).size());
    for (long long i = 0; i < n / 2; i++) {
      if (A[i] != A[n - i - 1]) return 0;
    }
    return 1;
  }
  template <typename T, typename T1>
  T amax(T &a, T1 b) {
    if (b > a) a = b;
    return a;
  }
  template <typename T, typename T1>
  T amin(T &a, T1 b) {
    if (b < a) a = b;
    return a;
  }
  void solve() {
    long long a, b;
    cin >> a >> b;
    if (a == b && a == 0)
      cout << 0 << '\n';
    else if (a == b and a != 0)
      cout << 1 << '\n';
    else if ((a + b) % 2 == 0)
      cout << 2 << '\n';
    else
      cout << -1 << '\n';
  }
  signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    long long tests = 1;
    cin >> tests;
    for (long long test = 1; test <= tests; test++) {
      solve();
    }
    return 0;
  }
  