#include <bits/stdc++.h>
using namespace std;
int main() {
  std::ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  cin >> n;
  while (n--) {
    long long x, y;
    cin >> x >> y;
    if (x == 0 && y == 0) {
      cout << "0\n";
    } else if (x == y) {
      cout << "1\n";
    } else if (abs(x - y) % 2 == 0) {
      cout << "2\n";
    } else {
      cout << "-1\n";
    }
  }
  return 0;
}

### Prompt

Please create a solution in CPP to the following problem:

<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). 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
  std::ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  cin >> n;
  while (n--) {
    long long x, y;
    cin >> x >> y;
    if (x == 0 && y == 0) {
      cout << "0\n";
    } else if (x == y) {
      cout << "1\n";
    } else if (abs(x - y) % 2 == 0) {
      cout << "2\n";
    } else {
      cout << "-1\n";
    }
  }
  return 0;
}

```