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:
import math
def pgcd(a,b):
    while (b != 0):
        r= a%b
        a=b
        b=r
    return a

a,b=map(int,input().split())
g= pgcd(a,b)
gDiv=set()
for i in range (1, int(math.sqrt(g))+1):
    if g%i==0:
        gDiv.add(i)
        gDiv.add(g//i)


ens=sorted(gDiv)
div=list(ens)
lenx = len(ens)
for i in range(int(input())):
    a, b = list(map(int, input().split(' ')))
    lo = 0
    hi = lenx-1
    work = -1
    while lo <= hi:
        mid = (lo + hi) // 2
        if div[mid] < a:
            lo = mid+1
        elif div[mid] > b:
            hi = mid - 1
        else:
            work = max(work, div[mid])
            lo = mid + 1
            
    print(work)