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:
#!/usr/bin/env python3

from math import sqrt
from bisect import bisect_right

# Calculate greatest common divisor of two numbers
# Ex: gcd(9, 27) -> gcd(18, 9) -> gcd(0, 9) -> 9
def gcd(a, b): 
	if a:
		return gcd(b % a, a) 
	else:
		return b

a, b = map(int, input().split())

d = gcd(a, b)

# Calculate all possible gcd values. Place lower end divisors below the sqrt in u and higher end in v
u, v = [1], [d]
for i in range(2, int(sqrt(d)) + 1):
    if d % i == 0:
        u.append(i)
        v.append(d // i)

# Combine into one ordered list
u += reversed(v)

# def bi_search(list_divs, val):
# 	lo = 0
# 	hi = len(list_divs) -1
# 	# print(list_divs)
# 	while lo < hi:
# 		# print(lo, hi)
# 		mid = (lo + hi)//2
# 		if list_divs[mid] < val:
# 			lo = mid + 1
# 		# elif list_divs[mid] == val:
# 		# 	return mid + 1
# 		else:
# 			high = mid
# 	# print(lo)
# 	if list_divs[lo] <= val:
# 		return lo
# 	return lo - 1




for i in range(int(input())):
    l, r = map(int, input().split())
    # This finds the largest value within the divisor list that is closest to r
    k = u[bisect_right(u, r) - 1]
    # k = u[bi_search(u, r) ]
    # print("k is:  ", k)
    # In the case that the value found is less than the lower bound, print -1
    # else print the value found
    print(-1 if l > k else k)