#include <bits/stdc++.h>
using namespace std;
using ll = long long;
void debug_out() { cerr << '\n'; }
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cerr << " " << to_string(H);
  debug_out(T...);
}
const ll MOD = 1000000007;
template <typename T>
inline void print_vector(vector<T> &a) {
  for (auto &x : a) cout << x << ' ';
  cout << '\n';
}
inline ll binary_pow(ll a, ll b) {
  ll res = 1;
  while (b > 0) {
    if (b & 1) res = res * a;
    a = a * a;
    b >>= 1;
  }
  return res;
}
inline ll mod_pow(ll x, ll y, ll m = MOD) {
  ll res = 1;
  x = x % m;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1) res = (res * x) % m;
    y = y >> 1;
    x = (x * x) % m;
  }
  return res;
}
inline ll mod_add(ll a, ll b, ll m = MOD) {
  a = a % m;
  b = b % m;
  return (((a + b) % m) + m) % m;
}
inline ll mod_mul(ll a, ll b, ll m = MOD) {
  a = a % m;
  b = b % m;
  return (((a * b) % m) + m) % m;
}
inline ll mod_sub(ll a, ll b, ll m = MOD) {
  a = a % m;
  b = b % m;
  return (((a - b) % m) + m) % m;
}
inline ll mminvprime(ll a, ll b) { return mod_pow(a, b - 2, b); }
inline ll mod_div(ll a, ll b, ll m = MOD) {
  a = a % m;
  b = b % m;
  return (mod_mul(a, mminvprime(b, m), m) + m) % m;
}
inline ll ceilf(ll x, ll y) { return x % y == 0 ? x / y : x / y + 1; }
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
void test_case() {
  ll c, d;
  cin >> c >> d;
  if (abs(c - d) & 1) {
    cout << "-1" << '\n';
    return;
  }
  if (c == d) {
    if (c == 0)
      cout << 0 << '\n';
    else
      cout << 1 << '\n';
  } else
    cout << 2 << '\n';
  return;
}
int main() {
  ios::sync_with_stdio(0);
  std::cin.tie(0);
  std::cout.tie(0);
  int t = 1;
  cin >> t;
  for (int i = 1; i <= t; i++) {
    test_case();
  }
  return 0;
}

### Prompt

Generate 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;
using ll = long long;
void debug_out() { cerr << '\n'; }
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cerr << " " << to_string(H);
  debug_out(T...);
}
const ll MOD = 1000000007;
template <typename T>
inline void print_vector(vector<T> &a) {
  for (auto &x : a) cout << x << ' ';
  cout << '\n';
}
inline ll binary_pow(ll a, ll b) {
  ll res = 1;
  while (b > 0) {
    if (b & 1) res = res * a;
    a = a * a;
    b >>= 1;
  }
  return res;
}
inline ll mod_pow(ll x, ll y, ll m = MOD) {
  ll res = 1;
  x = x % m;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1) res = (res * x) % m;
    y = y >> 1;
    x = (x * x) % m;
  }
  return res;
}
inline ll mod_add(ll a, ll b, ll m = MOD) {
  a = a % m;
  b = b % m;
  return (((a + b) % m) + m) % m;
}
inline ll mod_mul(ll a, ll b, ll m = MOD) {
  a = a % m;
  b = b % m;
  return (((a * b) % m) + m) % m;
}
inline ll mod_sub(ll a, ll b, ll m = MOD) {
  a = a % m;
  b = b % m;
  return (((a - b) % m) + m) % m;
}
inline ll mminvprime(ll a, ll b) { return mod_pow(a, b - 2, b); }
inline ll mod_div(ll a, ll b, ll m = MOD) {
  a = a % m;
  b = b % m;
  return (mod_mul(a, mminvprime(b, m), m) + m) % m;
}
inline ll ceilf(ll x, ll y) { return x % y == 0 ? x / y : x / y + 1; }
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
void test_case() {
  ll c, d;
  cin >> c >> d;
  if (abs(c - d) & 1) {
    cout << "-1" << '\n';
    return;
  }
  if (c == d) {
    if (c == 0)
      cout << 0 << '\n';
    else
      cout << 1 << '\n';
  } else
    cout << 2 << '\n';
  return;
}
int main() {
  ios::sync_with_stdio(0);
  std::cin.tie(0);
  std::cout.tie(0);
  int t = 1;
  cin >> t;
  for (int i = 1; i <= t; i++) {
    test_case();
  }
  return 0;
}

```