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;
  void debug_out() { cerr << '\n'; }
  template <typename Head, typename... Tail>
  void debug_out(Head H, Tail... T) {
    cerr << " " << to_string(H);
    debug_out(T...);
  }
  const ll MOD = 1000000007;
  template <typename T>
  inline void print_vector(vector<T> &a) {
    for (auto &x : a) cout << x << ' ';
    cout << '\n';
  }
  inline ll binary_pow(ll a, ll b) {
    ll res = 1;
    while (b > 0) {
      if (b & 1) res = res * a;
      a = a * a;
      b >>= 1;
    }
    return res;
  }
  inline ll mod_pow(ll x, ll y, ll m = MOD) {
    ll res = 1;
    x = x % m;
    if (x == 0) return 0;
    while (y > 0) {
      if (y & 1) res = (res * x) % m;
      y = y >> 1;
      x = (x * x) % m;
    }
    return res;
  }
  inline ll mod_add(ll a, ll b, ll m = MOD) {
    a = a % m;
    b = b % m;
    return (((a + b) % m) + m) % m;
  }
  inline ll mod_mul(ll a, ll b, ll m = MOD) {
    a = a % m;
    b = b % m;
    return (((a * b) % m) + m) % m;
  }
  inline ll mod_sub(ll a, ll b, ll m = MOD) {
    a = a % m;
    b = b % m;
    return (((a - b) % m) + m) % m;
  }
  inline ll mminvprime(ll a, ll b) { return mod_pow(a, b - 2, b); }
  inline ll mod_div(ll a, ll b, ll m = MOD) {
    a = a % m;
    b = b % m;
    return (mod_mul(a, mminvprime(b, m), m) + m) % m;
  }
  inline ll ceilf(ll x, ll y) { return x % y == 0 ? x / y : x / y + 1; }
  mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
  void test_case() {
    ll c, d;
    cin >> c >> d;
    if (abs(c - d) & 1) {
      cout << "-1" << '\n';
      return;
    }
    if (c == d) {
      if (c == 0)
        cout << 0 << '\n';
      else
        cout << 1 << '\n';
    } else
      cout << 2 << '\n';
    return;
  }
  int main() {
    ios::sync_with_stdio(0);
    std::cin.tie(0);
    std::cout.tie(0);
    int t = 1;
    cin >> t;
    for (int i = 1; i <= t; i++) {
      test_case();
    }
    return 0;
  }
  