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 JAVA):

Solution:
import java.util.*;

public class C67 {
	public static void main (String[] args) {
		Scanner in = new Scanner(System.in);
		int a = in.nextInt(), b = in.nextInt(), n = in.nextInt();
		
		//LinkedList<Integer> divisors = new LinkedList<Integer>();
		TreeSet<Integer> divisors = new TreeSet<Integer>();
		for (int i = 1; i * i <= a; i++)
			if (a % i == 0) {
				if (b % i == 0) divisors.add(i);
				if (b % (a / i) == 0) divisors.add(a / i);
			}
		for (int i = 1; i * i <= b; i++)
			if (b % i == 0) {
				if (a % i == 0) divisors.add(i);
				if (a % (b / i) == 0) divisors.add(b / i);
			}
		
		LinkedList<Integer> d = new LinkedList<Integer>(divisors);
		//Collections.sort(divisors);
		
		//System.out.println(Arrays.toString(d.toArray()));
		
		for (int i = 0; i < n; i++) {
			int l = in.nextInt(), h = in.nextInt();
			int left = 0, right = d.size() - 1, mid, result = -1;
			
			while (left <= right) {
				mid = (left + right) / 2;
				if (d.get(mid) >= l && d.get(mid) <= h) result = Math.max(result, d.get(mid));
				if (d.get(mid) > h) right = mid - 1; else left = mid + 1; 
			}
			
			System.out.println(result);
		}
	}
}