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.File;
import java.util.*;

public class problem75C {
	static int gcd, len;
	static List<Integer> factors;
	
	public static void main(String[] args) throws Exception {
		Scanner in = new Scanner(System.in);
		int a = in.nextInt();
		int b = in.nextInt();
		int n = in.nextInt();
		gcd = gcd(a,b);
		factors = new ArrayList<Integer>();
		
		for (int i = 1; i <= Math.sqrt(gcd); i++) {
			if (i*i==gcd) factors.add(i);
			else if (gcd%i == 0) {
				factors.add(i);
				factors.add(gcd/i);
			}
		}
		
		len = factors.size();
		Collections.sort(factors);
		
		for (int i = 0; i < n; i++) {
			int x = in.nextInt();
			int y = in.nextInt();
			System.out.println(query(x, y));
		}
	}
	
	static int gcd(int a, int b) {
	   if (b==0) return a;
	   return gcd(b,a%b);
	}
	
	static int query(int x, int y) {
		int a = 0;
		int b = len-1;
		int max = -1;
		
		while (a <= b) {
			int med = (a+b)/2;
			int m = factors.get(med);
			//System.out.println(m);
			if (x <= m && m <= y) { max = Math.max(max, m); }
			if (m == y) { return y; }
			if (m > y) { b = med-1; }
			if (m < y) { a = med+1; }
			
			if (a == b && med*2==a+b) break;
		}
		
		return max;
	}
}