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 sys

# sys.stdin = open('input.txt', 'r')
# sys.stdout = open('output.txt', 'w')

def gcd(num1, num2):
	if num2 == 0:
		return num1

	return gcd(num2, num1 % num2)


# def binary_s(left, right):
# 	while left <= right:
# 		mid = left + (right - left)//2

# 		if res1[mid] > right:
# 			right = mid - 1
# 		else 


a, b = map(int, input().split())
n = int(input())
k = gcd(a, b)
res1 = []

i = 1
while i * i <= k:
	if k % i == 0:
		res1.append(i)
		if i != k // i:
			res1.append(k // i)
	i += 1

for _ in range(n):
	l, r = map(int, input().split())
	ans = -1

	for d in res1:
		if l <= d <= r:
			ans = max(ans, d)

	print(ans)