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;
  void show(int ans[], int n) {
    for (int i = (0); i < (n); ++i) {
      cout << ans[i] << " ";
    }
    cout << "\n";
  }
  bool isPrime(int n) {
    for (int i = 2; i <= sqrt(n); ++i) {
      if (n % i == 0) {
        return false;
      }
    }
    return true;
  }
  int gcd(int a, int b) {
    while (a) {
      b %= a;
      a ^= b;
      b ^= a;
      a ^= b;
    }
    return b;
  }
  int tempa[10000];
  void mergesort(int* a, int n1, int n2) {
    int a1, a2, n12;
    int i, j;
    if (n1 < n2) {
      n12 = (n1 + n2) / 2;
      mergesort(a, n1, n12);
      mergesort(a, n12 + 1, n2);
      for (i = n1; i <= n2; i++) {
        tempa[i] = a[i];
      }
      a1 = n1;
      a2 = n12 + 1;
      for (i = n1; (a1 <= n12 && a2 <= n2); i++) {
        if (tempa[a1] <= tempa[a2]) {
          a[i] = tempa[a1];
          a1++;
        } else {
          a[i] = tempa[a2];
          a2++;
        }
      }
      if (a1 > n12) {
        for (j = a2; j <= n2; j++, i++) {
          a[i] = tempa[j];
        }
      } else {
        for (j = a1; j <= n12; j++, i++) {
          a[i] = tempa[j];
        }
      }
    }
  }
  bool sorted(string arr, int n) {
    for (int i = 0; i < n - 1; ++i) {
      if (arr[i + 1] != arr[i]) return false;
    }
    return true;
  }
  void bubbleSort(int* arr, int n) {
    for (int i = (0); i < (n); ++i) {
      for (int j = i; j < n; ++j) {
      }
    }
  }
  int solve() {
    int _test;
    cin >> _test;
    while (_test--) {
      int c, d, a = 0, b = 0;
      cin >> c >> d;
      if (c == d)
        cout << (c != 0) << "\n";
      else if (abs(c - d) & 1) {
        cout << (-1) << "\n";
      } else
        cout << (2) << "\n";
    }
    return 0;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    solve();
    return 0;
  }
  