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) {
    cout << name << " : " << arg1 << "\n";
  }
  template <typename Arg1, typename... Args>
  void __f(const char* names, Arg1&& arg1, Args&&... args) {
    const char* comma = strchr(names + 1, ',');
    cout.write(names, comma - names) << " : " << arg1 << " | ";
    __f(comma + 1, args...);
  }
  long long spf[1000000 + 2];
  vector<long long> primes;
  void seive() {
    long long i, j;
    long long prime[1000000];
    for (i = 1; i < 1000000; i++) {
      prime[1000000] = 1;
    }
    for (i = 4; i < 1000000; i += 2) {
      prime[i] = 0;
    }
    for (i = 3; i * i < 1000000; i += 2) {
      if (prime[i]) {
        for (j = i * i; j < 1000000; j += i) {
          prime[j] = 0;
        }
      }
    }
    for (i = 2; i < 1000000; i++) {
      if (prime[i]) {
        primes.emplace_back(i);
      }
    }
  }
  long long factor(long long x) {
    long long ans = 1;
    long long i;
    for (i = 0; i < primes.size() and primes[i] * primes[i] <= x; i++) {
      if (x % primes[i] == 0) {
        long long cnt = 0;
        while (x % primes[i] == 0) {
          cnt++;
          x /= primes[i];
        }
        if (cnt & 1) {
          ans *= primes[i];
        }
      }
    }
    if (x != 1) {
      ans *= x;
    }
    return ans;
  }
  vector<vector<long long>> dis(3002, vector<long long>(3002));
  int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    long long i, j, k, n, l, m, t;
    cin >> t;
    while (t--) {
      long long c, d;
      cin >> c >> d;
      if (c == d and c == 0) {
        cout << 0 << "\n";
      } else if (c == d) {
        cout << 1 << "\n";
      } else {
        long long dif = abs(c - d);
        if (dif % 2) {
          cout << -1 << "\n";
        } else {
          cout << 2 << "\n";
        }
      }
    }
  }
  