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:
a,b=map(int,input().split())
import math
g = math.gcd(a, b)
#  generate the divisor
divisor =[]
i=1
while i*i<g+1:
    if g%i==0:
        divisor.append(i),divisor.append(int(g/i))
    if i*i==g:
        divisor.append(i)
    i+=1
divisor.sort()
n=int(input())
s=len(divisor)-1
for i in range(n):
    ans=0
    L=0
    H=s
    l,h=map(int,input().split())
# print("len:divisor",len(divisor))
    while L<=H:
        m=(L+H)//2
        if divisor[m]<=h:
            ans=m
            L=m+1
        else:
            H=m-1
    if divisor[ans] < l:
        print(-1)
    else: print(divisor[ans])