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=tuple(map(int,input().split()))
n=int(input())
def gcd(a,b):
    if(a==0):
        return b
    else:
        return gcd(b%a,a)
g=gcd(a,b) 
l=[]
def divisors(n):
    global l
    i=2
    while(i<=math.sqrt(n)):
        if(n%i==0):
            l.append(i)
            if(i!=(n//i)):
                l.append(n//i)
        i+=1
divisors(g) 
 
l.append(1)
l=sorted(l) 
for w in range(n):
    t=0
    low,high=tuple(map(int,input().split()))
    if(g>=low and g<=high):
        print(g) 
    else:
        for i in reversed(range(len(l))):
            if(l[i]>=low and l[i]<=high):
                print(l[i])
                t=1
                break
        if(t!=1):
            print(-1)
#print(gcd(90,36))