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;
  const int IINF = 0x3f3f3f3f;
  const long long LINF = 0x3f3f3f3f3f3f3f3fLL;
  const double DINF = numeric_limits<double>::infinity();
  const double DNAN = numeric_limits<double>::quiet_NaN();
  const double EPS = 1e-9;
  const double PI = acos((double)-1.0);
  const int DX[] = {1, 0, -1, 0, 1, -1, 1, -1};
  const int DY[] = {0, 1, 0, -1, 1, -1, -1, 1};
  long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
  long long sqr(long long x) { return x * x; }
  long long sqr(int x) { return (long long)x * x; }
  double sqr(double x) { return x * x; }
  mt19937 mmtw(960172);
  long long rnd(long long x, long long y) {
    static uniform_int_distribution<long long> d;
    return d(mmtw) % (y - x + 1) + x;
  }
  template <typename T>
  T& updMin(T& a, T const& b) {
    if (b < a) a = b;
    return a;
  }
  template <typename T>
  T& updMax(T& a, T const& b) {
    if (a < b) a = b;
    return a;
  }
  long long divFloor(long long a, long long b) {
    if (b < 0) {
      a = -a;
      b = -b;
    }
    return a >= 0 ? a / b : (a - b + 1) / b;
  }
  long long divCeil(long long a, long long b) {
    if (b < 0) {
      a = -a;
      b = -b;
    }
    return a >= 0 ? (a + b - 1) / b : a / b;
  }
  long long divFloorS(long long a, long long b) {
    if (b < 0) {
      a = -a;
      b = -b;
    }
    return a >= 0 ? (a + b - 1) / b - 1 : a / b - 1;
  }
  long long divCeilS(long long a, long long b) {
    if (b < 0) {
      a = -a;
      b = -b;
    }
    return a >= 0 ? a / b + 1 : (a - b + 1) / b + 1;
  }
  template <typename K, typename V>
  V getOrDef(map<K, V> const& a, K const& k, V const& def = V()) {
    auto it = a.find(k);
    return it == a.end() ? def : it->second;
  }
  template <typename K, typename V>
  V getOrDef(unordered_map<K, V> const& a, K const& k, V const& def = V()) {
    auto it = a.find(k);
    return it == a.end() ? def : it->second;
  }
  template <typename A, typename B>
  istream& operator>>(istream& in, pair<A, B>& p) {
    return in >> p.first >> p.second;
  }
  template <typename A, typename B>
  ostream& operator<<(ostream& out, pair<A, B> const& p) {
    return out << p.first << " " << p.second;
  }
  template <typename T>
  istream& operator>>(istream& in, vector<T>& a) {
    for (T& x : a) in >> x;
    return in;
  }
  template <typename T>
  ostream& operator<<(ostream& out, vector<T> const& a) {
    bool f = true;
    for (T const& x : a) {
      if (!f) out << " ";
      f = false;
      out << x;
    }
    return out;
  }
  template <typename T>
  bool isZero(T const& x) {
    return x == T(0);
  }
  bool isZero(float x) { return abs(x) < EPS; }
  bool isZero(double x) { return abs(x) < EPS; }
  bool isZero(long double x) { return abs(x) < EPS; }
  template <typename T>
  int sgn(T const& x) {
    if (isZero(x)) return 0;
    return x > 0 ? 1 : -1;
  }
  int main() {
    ios::sync_with_stdio(false);
    int t;
    cin >> t;
    for (int i = 0; i < t; ++i) {
      long long a, b;
      cin >> a >> b;
      if ((a + b) % 2 != 0) {
        cout << "-1\n";
        continue;
      }
      if (a == 0 && b == 0) {
        cout << "0\n";
        continue;
      }
      if (a + b == 0 || a - b == 0) {
        cout << "1\n";
        continue;
      }
      cout << "2\n";
    }
    return 0;
  }
  