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:
from math import gcd

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

d = []

i = 1
while i * i <= g:
    if not g % i:
        d.append(i)
    i += 1

for i in range(len(d) - 1, -1, -1):
    if d[i] * d[i] != g:
        d.append(g // d[i])

for test in range(int(input())):
    l, r = map(int, input().split())
    L = 0
    R = len(d) - 1
    while L < R:
        middle = (L + R + 1) >> 1
        if d[middle] > r:
            R = middle - 1
        else:
            L = middle
    if l <= d[L] <= r:
        print(d[L])
    else:
        print(-1)

exit(0)