#include <bits/stdc++.h>
using namespace std;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cerr << name << " : " << arg1 << '\n';
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
const long long mxN = 2e5 + 100;
string s;
long long a[mxN];
vector<long long> v[mxN];
long long n, m, k, q;
long long dp[mxN];
long long ans;
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
void test_case() {
  long long a, b;
  cin >> a >> b;
  if (a < b) {
    swap(a, b);
  }
  if ((a + b) % 2 == 1) {
    cout << -1 << '\n';
  } else if (a == b) {
    if (a != 0) {
      a = 1;
    }
    cout << a << '\n';
  } else {
    cout << 2 << '\n';
  }
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t;
  t = 1;
  cin >> t;
  for (long long i = 1; i <= t; i++) {
    test_case();
  }
  return 0;
}

### Prompt

Your task is to create a Cpp solution 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;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cerr << name << " : " << arg1 << '\n';
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
const long long mxN = 2e5 + 100;
string s;
long long a[mxN];
vector<long long> v[mxN];
long long n, m, k, q;
long long dp[mxN];
long long ans;
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
void test_case() {
  long long a, b;
  cin >> a >> b;
  if (a < b) {
    swap(a, b);
  }
  if ((a + b) % 2 == 1) {
    cout << -1 << '\n';
  } else if (a == b) {
    if (a != 0) {
      a = 1;
    }
    cout << a << '\n';
  } else {
    cout << 2 << '\n';
  }
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t;
  t = 1;
  cin >> t;
  for (long long i = 1; i <= t; i++) {
    test_case();
  }
  return 0;
}

```