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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Collections;
import java.util.StringTokenizer;

public class ModifiedGCD {

	public static int gcd(int a, int b)
	{
		return b == 0 ? a : gcd(b, a%b);
	}
	
	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer(br.readLine());
		int n1 = Integer.parseInt(st.nextToken());
		int n2 = Integer.parseInt(st.nextToken());
		int n = gcd(n1, n2);
		int max = (int)Math.sqrt(n);
		ArrayList<Integer> divisors = new ArrayList<Integer>();
		for (int i = 1; i <= max; i++) {
			if(n % i == 0)
			{
				if(!divisors.contains(i))divisors.add(i);
				if(!divisors.contains(n/i))divisors.add(n/i);
			}
		}
		Collections.sort(divisors);
		int q = Integer.parseInt(br.readLine());
		for (int i = 0; i < q; i++) {
			st = new StringTokenizer(br.readLine());
			int s = Integer.parseInt(st.nextToken());
			int e = Integer.parseInt(st.nextToken());
			if(s == e)
			{
				if(n % s == 0)
					System.out.println(s);
				else
					System.out.println(-1);
				continue;
			}
			int pos = Collections.binarySearch(divisors, e);
			if(pos > 0)
			{
				System.out.println(e);
				continue;
			}
			else
			{
				pos = -(pos+1) - 1;
				if(divisors.get(pos) >= s)
				{
					System.out.println(divisors.get(pos));
				}
				else
				{
					System.out.println(-1);
				}
			}
		}
	}
}