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;
  long long z;
  long long to_dec(string s) {
    char rim[7] = {'I', 'V', 'X', 'L', 'C', 'D', 'M'};
    long long decc[7] = {1, 5, 10, 50, 100, 500, 1000};
    long long res = 0;
    for (long long i = 0; i < s.size(); i++) {
      long long tdig;
      for (long long j = 0; j < 7; j++) {
        if (rim[j] == s[i]) tdig = decc[j];
      }
      if (i < (long long)s.size() - 1 &&
          (s[i] == 'I' && s[i + 1] == 'V' || s[i] == 'I' && s[i + 1] == 'X' ||
           s[i] == 'X' && s[i + 1] == 'L' || s[i] == 'X' && s[i + 1] == 'C' ||
           s[i] == 'C' && s[i + 1] == 'D' || s[i] == 'C' && s[i + 1] == 'M')) {
        tdig *= (-1);
      }
      res += tdig;
    }
    return res;
  }
  vector<long long> decomp(long long n) {
    vector<long long> res;
    for (long long i = 2; i * i <= n; i++) {
      while (n % i == 0) {
        res.push_back(i);
        n /= i;
      }
    }
    if (n > 1) res.push_back(n);
    return res;
  }
  bool IsDigit(char a) {
    if (a <= '9' && a >= '0')
      return 1;
    else
      return 0;
  }
  bool IsLetter(char a) {
    if ((a <= 'z' && a >= 'a') || (a <= 'Z' && a >= 'A'))
      return 1;
    else
      return 0;
  }
  signed F(long long a) {
    z = 1;
    if (a == 0) return 1;
    for (long long i = 1; i <= a; i++) {
      z = z * i;
    }
    return z;
  }
  bool IsPrime(long long a) {
    if (a <= 0) return 0;
    if (a == 1) return 0;
    if (a == 2) return 1;
    for (long long i = 2; i * i <= a; i++) {
      if (a % i == 0) return 0;
    }
    return 1;
  }
  bool cmirt(long long n) {
    vector<long long> m;
    if (IsPrime(n)) return 0;
    long long l;
    m = decomp(n);
    long long sm1 = 0;
    z = n;
    while (z > 0) {
      sm1 += z % 10;
      z /= 10;
    }
    long long sm2 = 0;
    for (long long i = 0; i < m.size(); i++) {
      z = m[i];
      while (z > 0) {
        sm2 += z % 10;
        z /= 10;
      }
    }
    if (sm2 == sm1)
      return 1;
    else
      return 0;
  }
  vector<long long> srt(vector<long long>& v) {
    for (long long i = 0; i < v.size() - 1; i++) {
      for (long long j = i + 1; j < v.size(); j++) {
        if (v[i] > v[j]) swap(v[i], v[j]);
      }
    }
    return v;
  }
  long long kol_sosed(long long x, long long y, vector<string> v, char h,
                      long long n, long long m) {
    long long res;
    long long dx[8] = {-1, -1, 0, 1, 1, 1, 0, -1};
    long long dy[8] = {0, 1, 1, 1, 0, -1, -1, -1};
    for (long long i = 0; i < 8; i++) {
      long long nx = x + dx[i];
      long long ny = y + dy[i];
      if (nx < 0) nx += n;
      if (ny < 0) ny += m;
      nx = nx % n;
      ny = ny % m;
      res += (v[nx][ny] == h);
    }
    return res;
  }
  vector<long long> srtl(vector<long long> v, long long mx) {
    vector<long long> pockets(mx + 1);
    for (long long i = 0; i < (long long)v.size(); i++) {
      pockets[v[i]]++;
    }
    long long k = 0;
    for (long long i = 0; i < (long long)pockets.size(); i++) {
      while (pockets[i] > 0) {
        v[k] = i;
        pockets[i]--;
        k++;
      }
    }
    return v;
  }
  vector<long long> maxr(vector<long long> v, long long n) {
    long long mx = -1e18;
    vector<long long> maxr(n, -1e18);
    for (long long i = n - 1; i >= 0; i--) {
      mx = max(mx, v[i]);
      maxr[i] = mx;
    }
    return maxr;
  }
  vector<long long> maxl(vector<long long> v, long long n) {
    long long mx = -1e18;
    vector<long long> maxl(n, -1e18);
    for (long long i = 0; i < n; i++) {
      mx = max(mx, v[i]);
      maxl[i] = mx;
    }
    return maxl;
  }
  string in_chr(long long n) {
    string s = "";
    while (n > 0) {
      s += (char)(n % 10) + 48;
      n /= 10;
    }
    return s;
  }
  void solve() {
    long long a, b;
    cin >> a >> b;
    if (a < b) {
      z = a;
      a = b;
      b = z;
    }
    if ((a - b) % 2 == 1) {
      cout << -1 << endl;
      return;
    }
    if (a == b) {
      if (a == 0) {
        cout << 0 << endl;
      } else {
        cout << 1 << endl;
      }
    } else {
      cout << 2 << endl;
    }
  }
  signed main() {
    long long u;
    cin >> u;
    while (u--) {
      solve();
    }
  }
  