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 PYTHON):

Solution:
'''
Created on 2011-5-3

@author: Sirius
'''
import math


def fac(n):
    L = [] 
    i = 1
    while i * i <= n:
        if n % i == 0:
            L.append(i)
            if i * i != n :
                L.append(n / i)
        i = i + 1
    return L

n, m = map(int, raw_input().split())
L = list( set(fac(n)) & set(fac(m)) )
L.sort()

k = int(raw_input())
for var in xrange(k) :
    Low, High = map(int, raw_input().split())
    l = 0
    r = len(L) - 1
    while l <= r :
        mid = ( l + r ) // 2
        if L[mid] <= High:
            l = mid + 1
        else:
            r = mid - 1
    if r < 0 or L[r] > High or L[r] < Low:
        print -1
    else :
        print L[r]