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

def divisors(n):
    div = []

    for i in range(1, int(n ** (1 / 2)) + 1):
        if n % i == 0:
            div.append(i)
            if i * i != n:
                div.append(n // i)

    return sorted(div)

d = sorted(set(divisors(a)).intersection(divisors(b)))

for _ in range(int(input())):
	l, r = map(int, input().split())
	from bisect import bisect_left
	t = bisect_left(d, r)
	print(d[t] if t < len(d) and d[t] == r else d[t - 1] if d[t - 1] >= l else -1)