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;
  long long a[5000007], b[5000007], c[5000007];
  long long mod = 1e9 + 7;
  long long power(long long x, long long y, long long p) {
    long long res = 1;
    x = x % p;
    if (x == 0) return 0;
    while (y > 0) {
      if (y & 1) res = (res * x) % p;
      y = y >> 1;
      x = (x * x) % p;
    }
    return res;
  }
  void SieveOfEratosthenes(long long n) {
    bool prime[n + 1];
    long long j = 0;
    memset(prime, true, sizeof(prime));
    for (long long p = 2; p * p <= n; p++) {
      if (prime[p] == true) {
        for (long long i = p * p; i <= n; i += p) prime[i] = false;
      }
    }
    for (long long p = 2; p <= n; p++)
      if (prime[p]) prime[j++] = p;
  }
  bool isPrime(long long n) {
    if (n <= 1) return false;
    for (long long i = 2; i < n; i++)
      if (n % i == 0) return false;
    return true;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    long long t = 1;
    cin >> t;
    while (t--) {
      long long n, i, j, m, k, x, flag = 1;
      std::map<long long, long long> mp1;
      cin >> n >> m;
      if (abs(n - m) % 2)
        cout << "-1\n";
      else {
        if (n == m && n == 0)
          cout << "0\n";
        else if (n == m)
          cout << "1\n";
        else
          cout << "2\n";
      }
    }
    return 0;
  }
  