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
def __gcd(a,b):
    if b==0:
        return a
    else:
        return __gcd(b,a%b)
def main():
    a,b=map(int,stdin.readline().split())
    gcd=__gcd(a,b)
    i=1
    factor=[]
    while i*i<=gcd:
        if gcd%i==0:
            factor.append(i)
            if i*i!=gcd:
                factor.append(gcd//i)
        i+=1
    q=int(stdin.readline())
    for i in range(q):
        res=-1
        l,r=map(int,stdin.readline().split())
        for j in range(len(factor)):
            if factor[j]>=l and factor[j]<=r:
                res=max(res,factor[j])
        stdout.write(str(res)+"\n")
if __name__=="__main__":
    main()