#include <bits/stdc++.h>
using namespace std;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cout << name << " : " << arg1 << "\n";
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cout.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
long long spf[1000000 + 2];
vector<long long> primes;
void seive() {
  long long i, j;
  long long prime[1000000];
  for (i = 1; i < 1000000; i++) {
    prime[1000000] = 1;
  }
  for (i = 4; i < 1000000; i += 2) {
    prime[i] = 0;
  }
  for (i = 3; i * i < 1000000; i += 2) {
    if (prime[i]) {
      for (j = i * i; j < 1000000; j += i) {
        prime[j] = 0;
      }
    }
  }
  for (i = 2; i < 1000000; i++) {
    if (prime[i]) {
      primes.emplace_back(i);
    }
  }
}
long long factor(long long x) {
  long long ans = 1;
  long long i;
  for (i = 0; i < primes.size() and primes[i] * primes[i] <= x; i++) {
    if (x % primes[i] == 0) {
      long long cnt = 0;
      while (x % primes[i] == 0) {
        cnt++;
        x /= primes[i];
      }
      if (cnt & 1) {
        ans *= primes[i];
      }
    }
  }
  if (x != 1) {
    ans *= x;
  }
  return ans;
}
vector<vector<long long>> dis(3002, vector<long long>(3002));
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long i, j, k, n, l, m, t;
  cin >> t;
  while (t--) {
    long long c, d;
    cin >> c >> d;
    if (c == d and c == 0) {
      cout << 0 << "\n";
    } else if (c == d) {
      cout << 1 << "\n";
    } else {
      long long dif = abs(c - d);
      if (dif % 2) {
        cout << -1 << "\n";
      } else {
        cout << 2 << "\n";
      }
    }
  }
}

### Prompt

Your challenge is to write 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) {
  cout << name << " : " << arg1 << "\n";
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cout.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
long long spf[1000000 + 2];
vector<long long> primes;
void seive() {
  long long i, j;
  long long prime[1000000];
  for (i = 1; i < 1000000; i++) {
    prime[1000000] = 1;
  }
  for (i = 4; i < 1000000; i += 2) {
    prime[i] = 0;
  }
  for (i = 3; i * i < 1000000; i += 2) {
    if (prime[i]) {
      for (j = i * i; j < 1000000; j += i) {
        prime[j] = 0;
      }
    }
  }
  for (i = 2; i < 1000000; i++) {
    if (prime[i]) {
      primes.emplace_back(i);
    }
  }
}
long long factor(long long x) {
  long long ans = 1;
  long long i;
  for (i = 0; i < primes.size() and primes[i] * primes[i] <= x; i++) {
    if (x % primes[i] == 0) {
      long long cnt = 0;
      while (x % primes[i] == 0) {
        cnt++;
        x /= primes[i];
      }
      if (cnt & 1) {
        ans *= primes[i];
      }
    }
  }
  if (x != 1) {
    ans *= x;
  }
  return ans;
}
vector<vector<long long>> dis(3002, vector<long long>(3002));
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long i, j, k, n, l, m, t;
  cin >> t;
  while (t--) {
    long long c, d;
    cin >> c >> d;
    if (c == d and c == 0) {
      cout << 0 << "\n";
    } else if (c == d) {
      cout << 1 << "\n";
    } else {
      long long dif = abs(c - d);
      if (dif % 2) {
        cout << -1 << "\n";
      } else {
        cout << 2 << "\n";
      }
    }
  }
}

```