#include <bits/stdc++.h>
using namespace std;
inline char gc() {
  static char buf[100000], *p1 = buf, *p2 = buf;
  return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2)
             ? EOF
             : *p1++;
}
inline long long read() {
  char c = getchar();
  long long su = 0, f = 1;
  for (; c < '0' || c > '9'; c = getchar())
    if (c == '-') f = -1;
  for (; c >= '0' && c <= '9'; c = getchar()) su = su * 10 + c - '0';
  return su * f;
}
inline void write(long long x) {
  if (x < 0) {
    putchar('-');
    write(-x);
    return;
  }
  if (x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}
inline void writesp(long long x) { write(x), putchar(' '); }
inline void writeln(long long x) {
  write(x);
  putchar('\n');
}
int T, x, y, n, m;
signed main() {
  T = read();
  while (T--) {
    x = 0, y = 0, n = read(), m = read();
    if ((x + y) % 2 != (n + m) % 2)
      puts("-1");
    else {
      if (x == n && y == m)
        puts("0");
      else if (n - x == m - y || x + y == n + m)
        puts("1");
      else
        puts("2");
    }
  }
}

### 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;
inline char gc() {
  static char buf[100000], *p1 = buf, *p2 = buf;
  return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2)
             ? EOF
             : *p1++;
}
inline long long read() {
  char c = getchar();
  long long su = 0, f = 1;
  for (; c < '0' || c > '9'; c = getchar())
    if (c == '-') f = -1;
  for (; c >= '0' && c <= '9'; c = getchar()) su = su * 10 + c - '0';
  return su * f;
}
inline void write(long long x) {
  if (x < 0) {
    putchar('-');
    write(-x);
    return;
  }
  if (x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}
inline void writesp(long long x) { write(x), putchar(' '); }
inline void writeln(long long x) {
  write(x);
  putchar('\n');
}
int T, x, y, n, m;
signed main() {
  T = read();
  while (T--) {
    x = 0, y = 0, n = read(), m = read();
    if ((x + y) % 2 != (n + m) % 2)
      puts("-1");
    else {
      if (x == n && y == m)
        puts("0");
      else if (n - x == m - y || x + y == n + m)
        puts("1");
      else
        puts("2");
    }
  }
}

```