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:
#input template
from sys import stdin, stdout
cin = stdin.readline
cout = stdout.write
mp = lambda: list(map(int, cin().split()))

def chars(): #function for taking string input as character array since string in python is immutable
    s = cin()
    return(list(s[:len(s) - 1]))

#print list	
def pl(a):	
	for val in a:
		cout(val + '\n')
		
#main

def binSearch(lo, hi):
	high = len(l) - 1
	low = 0
	ans = -1
	while low <= high:
		mid = (low+high)//2
		#print(mid, l[mid])
		if hi>=l[mid] and l[mid]>= lo:
			ans = l[mid]
			#print(ans, 120)
			low = mid+1
		elif l[mid] < lo:
			low = mid+1
		elif l[mid] > hi:
			high = mid-1
	return ans


a, b = sorted(mp())
n, =  mp()

s = set()
for i in range(1, int(a**.5)+1):
	if not a%i and not b%i:
		s.add(i)
	if not a%(a//i) and not b%(a//i):
		s.add(a//i)

l = sorted(s)
#print(l)			
for _ in range(n):
	lo, hi = mp()
	
	ans = binSearch(lo, hi)
	cout(str(ans) + '\n')