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 sys import stdin
from bisect import bisect


def divisors(n):
    ans = set()
    i = 1
    while i * i <= n:
        if n % i == 0:
            ans.add(i)
        i += 1

    return ans | {n // e for e in ans}


test = stdin.readlines()
a, b = map(int, test[0].split())
n = int(test[1])

cdiv = sorted(divisors(a) & divisors(b))
m = len(cdiv)
out = []
for i in range(2, 2 + n):
    l, r = map(int, test[i].split())
    idx = bisect(cdiv, r) - 1
    out.append(cdiv[idx] if 0 <= idx < m and l <= cdiv[idx] <= r else -1)

print('\n'.join(map(str, out)))