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.io.*;
import java.util.*;

public class Main {
//	static Scanner in;
	static PrintWriter out;
	static StreamTokenizer in; static int next() throws Exception {in.nextToken(); return (int) in.nval;}

	public static void main(String[] args) throws Exception {
//		in = new Scanner(System.in);
		out = new PrintWriter(System.out);
		in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));

		int a = next();
		int b = next();

		TreeSet<Integer> set = new TreeSet<Integer>();

		for (int i = 1; i*i <= a; i++) if (a % i == 0) {
			if (b % i == 0) set.add(i);
			if (b % (a / i) == 0) set.add(a/i);
		}

		int k = set.size();
		int[] d = new int[k];
		k = 0;
		for (int x : set) d[k++] = x;

		int n = next();
		for (int i = 0; i < n; i++) {
			int L = next();
			int R = next();
			int max = -1;
			for (int x : d) if (L <= x && x <= R) max = Math.max(max, x);
			out.println(max);
		}

		out.close();
	}
}