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:
import sys
from math import sqrt
def mcd(p,q):
    if p%q==0:
        return q
    return mcd(q, p%q)

def divMcd(n):
    div=[]
    for i in range (1,int(sqrt(n))+1):
        if n % i==0:
            div.append(i)
            div.append(n//i)
    return div

a,b=map(int, sys.stdin.readline().split())
mcd_=mcd(a,b)
divisores=divMcd(mcd_)

divisores=list(set(divisores))
divisores.sort()
#print(divisores)
q=int(sys.stdin.readline())


for _ in range (q):
    l,h=map(int, sys.stdin.readline().split())
    max_=-1
    for i in range (len(divisores)):
        if divisores[i] >= l and divisores[i]<=h:
            max_=max(max_,divisores[i])
    print(max_)