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 math
a,b=map(int,input().split())
g=math.gcd(a,b)
#print(g)
l1=[]
for i in range(1,int(math.sqrt(g))+1):
    if g%i==0:
        l1.append(i)
        if i*i!=g:
            l1.append(g//i)
#print(l1)
l1.sort()
n=int(input())
for i in range(n):
    l,r=map(int,input().split())
    res=-1
    x=0
    h=len(l1)-1
    while x<=h:
        m=(x+h)//2
        if l1[m]>r:
            h=m-1
        elif l1[m]<l:
            x=m+1
        else:
            res=max(res,l1[m])
            x=m+1
    print(res)