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 sys import stdin
from bisect import bisect_left, bisect_right


def gcd(a, b):
    if a > b:
        a, b = b, a
    while a:
        a, b = b % a, a
    return b


a, b = map(int, stdin.readline().split())

g = gcd(a,b)
gfactor = [1,g]
p = 2
while p*p<=g:
    if g%p==0:
        gfactor.append(p)
        if g//p != p:
            gfactor.append(g//p)
    p += 1
gfactor = sorted(gfactor)

n = int(stdin.readline().strip())
for i in range(n):
    low, high = map(int, stdin.readline().split())
    ridx = bisect_right(gfactor, high) - 1
    if gfactor[ridx]>=low:
        print(gfactor[ridx])
    else:
        print(-1)