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(long long t) { cerr << t; }
  void show(int t) { cerr << t; }
  void show(string t) { cerr << t; }
  void show(char t) { cerr << t; }
  void show(long double t) { cerr << t; }
  void show(double t) { cerr << t; }
  void show(unsigned long long t) { cerr << t; }
  template <class Y, class V>
  void show(pair<Y, V> p);
  template <class Y>
  void show(vector<Y> v);
  template <class Y>
  void show(set<Y> v);
  template <class Y>
  void show(multiset<Y> v);
  template <class Y, class V>
  void show(map<Y, V> v);
  template <class Y, class V>
  void show(pair<Y, V> p) {
    cerr << "{";
    show(p.first);
    cerr << ",";
    show(p.second);
    cerr << "}";
  }
  template <class Y>
  void show(set<Y> v) {
    cerr << "[ ";
    for (Y i : v) {
      show(i);
      cerr << " ";
    }
    cerr << "]";
  }
  template <class Y>
  void show(vector<Y> v) {
    cerr << "[ ";
    for (Y i : v) {
      show(i);
      cerr << " ";
    }
    cerr << "]";
  }
  template <class Y, class V>
  void show(map<Y, V> v) {
    cerr << "[ ";
    for (auto i : v) {
      show(i);
      cerr << " ";
    }
    cerr << "]";
  }
  template <class Y>
  void show(multiset<Y> v) {
    cerr << "[ ";
    for (Y i : v) {
      show(i);
      cerr << " ";
    }
    cerr << "]";
  }
  template <typename Head, typename... Tail>
  void show(Head H, Tail... T) {
    cerr << H << ' ';
    show(T...);
  }
  const long long inf = 1e18;
  const long long mod = 1e9 + 7;
  const long long N = 2e5 + 1;
  bool ispalin(string s) {
    long long len = s.length();
    for (long long i = 0; i < len; i++) {
      if (s[i] != s[len - i - 1]) {
        return false;
      }
    }
    return true;
  }
  long long add(long long a, long long b, long long m) {
    long long res = (a % m + b % m) % m;
    if (res < 0) {
      res += m;
    }
    return res;
  }
  long long mul(long long a, long long b, long long m) {
    long long res = (a % m * b % m) % m;
    if (res < 0) {
      res += m;
    }
    return res;
  }
  long long power(long long a, long long b, long long m) {
    long long res = 1;
    while (b > 0) {
      if (b & 1) {
        res = mul(res, a, m);
      }
      a = mul(a, a, m);
      b >>= 1;
    }
    return res % m;
  }
  long long binpow(long long a, long long b) {
    long long res = 1;
    while (b > 0) {
      if (b & 1) {
        res = (res * a);
      }
      a = (a * a);
      b >>= 1;
    }
    return res;
  }
  long long binpow(long long a, long long b, long long m) {
    long long res = 1;
    while (b > 0) {
      if (b & 1) {
        res = (res * a) % m;
      }
      a = (a * a) % m;
      b >>= 1;
    }
    return res % m;
  }
  bool isprime(long long n) {
    if (n <= 1) {
      return false;
    }
    if (n <= 3) {
      return true;
    }
    if (n % 2 == 0 || n % 3 == 0) {
      return false;
    }
    for (long long i = 5; i * i <= n; i = i + 6) {
      if (n % i == 0 || n % (i + 2) == 0) {
        return false;
      }
    }
    return true;
  }
  long long gcd(long long a, long long b) {
    if (b == 0) {
      return a;
    } else {
      return gcd(b, a % b);
    }
  }
  long long lcm(long long a, long long b) {
    long long l = (a * b) / gcd(a, b);
    return l;
  }
  long long CEIL(long long a, long long b) { return (a / b + (a % b != 0)); }
  long long inv(long long a, long long m) { return power(a, m - 2, m); }
  long long C(long long n, long long r) {
    if (n < r) return 0;
    if (n == r) return 1;
    if (r > n - r) r = n - r;
    long long ans = 1;
    for (long long i = 1; i <= r; i++) ans *= n - r + i, ans /= i;
    return ans;
  }
  bool isPowerOfTwo(long long x) { return x && (!(x & (x - 1))); }
  bool isPerfectSq(long long x) {
    if (x >= 0) {
      long long sr = sqrt(x);
      return (sr * sr == x);
    }
    return false;
  }
  long long fact(long long n, long long m) {
    long long f = 1;
    for (long long i = 1; i <= n; i++) f = (f * i) % m;
    return f % m;
  }
  long long fact_pow(long long n, long long k) {
    long long res = 0;
    while (n) {
      n /= k;
      res += n;
    }
    return res;
  }
  void solve() {
    long long c, d;
    cin >> c >> d;
    if (c < d) swap(c, d);
    if ((c - d) % 2)
      cout << "-1" << '\n';
    else if (c == 0 && d == 0)
      cout << "0" << '\n';
    else if (c == d)
      cout << "1" << '\n';
    else
      cout << "2" << '\n';
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int t = 1;
    cin >> t;
    while (t--) solve();
    return 0;
  }
  