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;
  using ll = long long;
  using ld = long double;
  using db = double;
  using str = string;
  using pi = pair<int, int>;
  using pl = pair<ll, ll>;
  using pd = pair<db, db>;
  using vi = vector<int>;
  using vb = vector<bool>;
  using vl = vector<ll>;
  using vd = vector<db>;
  using vs = vector<str>;
  using vpi = vector<pi>;
  using vpl = vector<pl>;
  using vpd = vector<pd>;
  template <class T>
  ll lwb(vector<T>& a, const T& b) {
    return lower_bound(a.begin(), a.end(), b) - a.begin();
  }
  template <class T>
  ll upb(vector<T>& a, const T& b) {
    return upper_bound(a.begin(), a.end(), b) - a.begin();
  }
  void setIn(str second) { freopen(second.c_str(), "r", stdin); }
  void setOut(str second) { freopen(second.c_str(), "w", stdout); }
  void setIO(str second) {
    setIn(second + ".inp");
    setOut(second + ".out");
  }
  const int MOD = 1e9 + 7;
  const int MX = 2e5 + 5;
  const ll INF = 1e18;
  const db PI = acos((db)-1);
  const int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
  template <class T>
  using pqg = priority_queue<T, vector<T>, greater<T>>;
  ll pow_mod(ll a, ll b, ll m) {
    ll res = 1;
    while (b) {
      res = res * (b & 1 ? a : 1) % m;
      a = a * a % m;
      b >>= 1;
    }
    return res;
  }
  ll inv(ll a, ll m) { return pow_mod(a, m - 2, m); }
  ll nxt(void) {
    ll x;
    cin >> x;
    return x;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    ll TC = nxt();
    while (TC--) {
      ll c, d;
      cin >> c >> d;
      if (abs(c - d) % 2) {
        cout << "-1\n";
        continue;
      }
      cout << (c != 0 || d != 0) + (c != d) << '\n';
    }
    return 0;
  }
  