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;
  using abu_bakr = int;
  long long mull(long long a, long long b, long long m) {
    return ((a % m) * (b % m)) % m;
  }
  long long gcd(long long a, long long b) {
    if (a < b) swap(a, b);
    while (a != 0 && b != 0) {
      long long r = a % b;
      a = b;
      b = r;
    }
    return a;
  }
  long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
  unsigned long long nCr(int n, int r) {
    if (r > n) return 0;
    r = max(r, n - r);
    unsigned long long ans = 1, div = 1, i = r + 1;
    while (i <= n) {
      ans *= i;
      i++;
      ans /= div;
      div++;
    }
    return ans;
  }
  void solve() {
    int a, b;
    cin >> a >> b;
    if (!a && !b)
      cout << 0;
    else if (abs(a - b) == 0)
      cout << 1;
    else if ((a + b) % 2 == 1)
      cout << -1;
    else
      cout << 2;
  }
  int main() {
    cin.tie(0)->sync_with_stdio(0);
    cin.exceptions(cin.failbit);
    int t;
    cin >> t;
    while (t--) {
      solve();
      cout << '\n';
    }
    return 0;
  }
  