Problem: Well, here is another math class task. In mathematics, GCD is the greatest common divisor, and it's an easy task to calculate the GCD between two positive integers.

A common divisor for two positive numbers is a number which both numbers are divisible by.

But your teacher wants to give you a harder task, in this task you have to find the greatest common divisor d between two integers a and b that is in a given range from low to high (inclusive), i.e. low ≤ d ≤ high. It is possible that there is no common divisor in the given range.

You will be given the two integers a and b, then n queries. Each query is a range from low to high and you have to answer each query.

Input

The first line contains two integers a and b, the two integers as described above (1 ≤ a, b ≤ 109). The second line contains one integer n, the number of queries (1 ≤ n ≤ 104). Then n lines follow, each line contains one query consisting of two integers, low and high (1 ≤ low ≤ high ≤ 109).

Output

Print n lines. The i-th of them should contain the result of the i-th query in the input. If there is no common divisor in the given range for any query, you should print -1 as a result for this query.

Examples

Input

9 27
3
1 5
10 11
9 11


Output

3
-1
9

Solution (in PYTHON3):

Solution:
from sys import stdin, stdout
from math import gcd, sqrt
from array import array

def getin(type):
    return type(stdin.readline().strip())

def getrin(type):
    return [type(x) for x in stdin.readline().strip().split()]

def getout(object, type):
    return stdout.write(str(type(object)) + '\n')

a, b = getrin(int)
JoJo = gcd(a, b)
factors = array('i', [])
i = 1
sJoJo = int(sqrt(JoJo)) + 1
while i <= sJoJo:
    if JoJo % i == 0:
        factors.append(i)
        if JoJo / i != i:
            factors.append(int(JoJo / i))
    i += 1
n = getin(int)
i = 0
while i < n:
    low, high = getrin(int)
    Max = -1
    for factor in factors:
        if low <= factor <= high:
            Max = max(Max, factor)
    getout(Max, int)
    i += 1