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 math import sqrt
 
def gcd(a,b):
	if b == 0:
		return a
	return gcd(b, a%b)
 
def binary_search(div, num):
	start, end = 0, len(div)
	while start < end:
		mid = int((end-start)/2)+start
		if div[mid] > num:
			end = mid
		else:
			start = mid + 1
	return div[start - 1]
 
a,b = map(int,input().split())
 
curr_gcd = gcd(a,b)
div = []
for i in range(1,int(sqrt(curr_gcd)) + 1):
	if curr_gcd % i == 0:
		div.append(i)
		if curr_gcd / i != i:
			div.append(int(curr_gcd/i))
div.sort()
n = int(input())
for i in range(n):
	low, hi = map(int, input().split())
	res=binary_search(div,hi)
	print(-1 if low>res else res)