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 PYTHON):

Solution:
def gcd(a,b):
    if a%b!=0:
        x=gcd(b,a%b)
    else:
        x=b
    return x
def calc(a,b,fact,fact1):
    i=-1
    for x in xrange(len(fact1)):
        if fact1[x]<=b and fact1[x]>=a:
            i=fact1[x]
            break
    if i!=-1: return i
    for x in range(len(fact)-1,-1,-1):
        if fact[x]<=b and fact[x]>=a:
            i=fact[x]
            break
    return i    
a,b=[int(x) for x in raw_input().split()]
c=int(raw_input())
n=gcd(a,b)
fact=[]
fact1=[]
for x in xrange(1,int(n**0.5)+1):
    if n%x==0:
        fact.append(x)
        fact1.append(n/x) 
for x in xrange(c):
    d,e=[int(x) for x in raw_input().split()]
    print(calc(d,e,fact,fact1))