human
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). 
# SHR ... What is 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.
i GANESHA    author: Kunal Verma #

import os
import sys
from collections import Counter

from itertools import *

from io import BytesIO, IOBase

def lcm(a, b):
    return (a * b) // gcd(a, b)

'''
    mod = 10 ** 9 + 7
    fac = [1]
    for i in range(1, 2 * 10 ** 5 + 1):
        fac.append((fac[-1] * i) % mod)
    fac_in = [pow(fac[-1], mod - 2, mod)]
    for i in range(2 * 10 ** 5, 0, -1):
        fac_in.append((fac_in[-1] * i) % mod)
    fac_in.reverse()
    def comb(a, b):
        if a < b:
            return 0
        return (fac[a] * fac_in[b] * fac_in[a - b]) % mod
'''

# MAXN = 10000004
# spf = [0 for i in range(MAXN)]
# adj = [[] for i in range(MAXN)]
def sieve():
    global spf, adj, MAXN
    spf[1] = 1
    for i in range(2, MAXN):
        spf[i] = i

    for i in range(2, MAXN):
        if i * i > MAXN:
            break
        if (spf[i] == i):
            for j in range(i * i, MAXN, i):
                if (spf[j] == j):
                    spf[j] = i

def getdistinctFactorization(n):
    global adj, spf, MAXN
    for i in range(1, n + 1):
        index = 1
        x = i
        if (x != 1):
            adj[i].append(spf[x])
        x = x // spf[x]
        while (x != 1):
            if (adj[i][index - 1] != spf[x]):
                adj[i].append(spf[x])
                index += 1
            x = x // spf[x]

def printDivisors(n):
    i = 2
    z = [1, n]
    while i <= sqrt(n):
        if (n % i == 0):
            if (n / i == i):
                z.append(i)
            else:
                z.append(i)
                z.append(n // i)
        i = i + 1
    return z

def create(n, x, f):
    pq = len(bin(n)[2:])
    if f == 0:
        tt = min
    else:
        tt = max
    dp = [[inf] * n for _ in range(pq)]
    dp[0] = x
    for i in range(1, pq):
        for j in range(n - (1 << i) + 1):
            dp[i][j] = tt(dp[i - 1][j], dp[i - 1][j + (1 << (i - 1))])
    return dp

def enquiry(l, r, dp, f):
    if l > r:
        return inf if not f else -inf
    if f == 1:
        tt = max
   ...