human
Can you solve the below in CPP?

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). 
 Write me a Python program 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?
assistant
#Here is Python program 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.

#include <bits/stdc++.h>
using namespace std;
const int IINF = 0x3f3f3f3f;
const long long LINF = 0x3f3f3f3f3f3f3f3fLL;
const double DINF = numeric_limits<double>::infinity();
const double DNAN = numeric_limits<double>::quiet_NaN();
const double EPS = 1e-9;
const double PI = acos((double)-1.0);
const int DX[] = {1, 0, -1, 0, 1, -1, 1, -1};
const int DY[] = {0, 1, 0, -1, 1, -1, -1, 1};
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long sqr(long long x) { return x * x; }
long long sqr(int x) { return (long long)x * x; }
double sqr(double x) { return x * x; }
mt19937 mmtw(960172);
long long rnd(long long x, long long y) {
  static uniform_int_distribution<long long> d;
  return d(mmtw) % (y - x + 1) + x;
}
template <typename T>
T& updMin(T& a, T const& b) {
  if (b < a) a = b;
  return a;
}
template <typename T>
T& updMax(T& a, T const& b) {
  if (a < b) a = b;
  return a;
}
long long divFloor(long long a, long long b) {
  if (b < 0) {
    a = -a;
    b = -b;
  }
  return a >= 0 ? a / b : (a - b + 1) / b;
}
long long divCeil(long long a, long long b) {
  if (b < 0) {
    a = -a;
    b = -b;
  }
  return a >= 0 ? (a + b - 1) / b : a / b;
}
long long divFloorS(long long a, long long b) {
  if (b < 0) {
    a = -a;
    b = -b;
  }
  return a >= 0 ? (a + b - 1) / b - 1 : a / b - 1;
}
long long divCeilS(long long a, long long b) {
  if (b < 0) {
    a = -a;
    b = -b;
  }
  return a >= 0 ? a / b + 1 : (a - b + 1) / b + 1;
}
template <typename K, typename V>
V getOrDef(map<K, V> const& a, K const& k, V const& def = V()) {
  auto it = a.find(k);
  return it == a.end() ? def : it->second;
}
template <typename K, typename V>
V getOrDef(unordered_map<K, V> const& a, K const& k, V const& def = V()) {
  auto it = a.find(k);
  return it == a.end() ? def : it->second;
}
template <typename A, typename B>
istream& operator>>(istream& in, pair<A, B>& p) {
  return in >> p.first >> p.second;
}
template <typename A, typename B>
ostream& operator<<(ostream& out, pair<A, B> const& p) {
  return out...