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.io.StreamTokenizer;
import java.util.Arrays;

public class C75 {

    public static void main(String[] args) throws IOException{
        StreamTokenizer sc = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
        sc.nextToken();
        int a = (int) sc.nval;
        sc.nextToken();
        int b = (int) sc.nval;
        int[] a_bol = new int[100000], b_bol = new int[100000];
        int[] a_b_bol = new int[100000];
        int ind_a = 0, ind_b = 0;
        for (int i = 1; i*i <= a; i++) {
            if (a % i == 0) {
                a_bol[++ind_a] = i;
                a_bol[++ind_a] = a / i;
            }
        }
        for (int i = 1; i*i <= b; i++) {
            if (b % i == 0) {
                b_bol[++ind_b] = i;
                b_bol[++ind_b] = b / i;
            }
        }
        Arrays.sort(a_bol, 1, ind_a+1);
        int ind_a_b = 0;
        for (int i = 1; i <= ind_b; i++) {
            if (Arrays.binarySearch(a_bol,1,ind_a+1, b_bol[i]) > 0) {
                a_b_bol[++ind_a_b] = b_bol[i];
            }
        }
        Arrays.sort(a_b_bol, 1, ind_a_b+1);
        sc.nextToken();
        int m = (int) sc.nval;
        for (int i = 1; i <= m; i++) {
            sc.nextToken();
            a = (int) sc.nval;
            sc.nextToken();
            b = (int) sc.nval;
            int l = Arrays.binarySearch(a_b_bol,1,ind_a_b+1, a);
            int r = Arrays.binarySearch(a_b_bol,1,ind_a_b+1, b);
//          System.out.println("l = "+l+" r = "+r);
            if (r > 0)
                System.out.println(a_b_bol[r]);
            else {
                if (Math.abs(l) < Math.abs(r))
                    System.out.println(a_b_bol[Math.abs(r)-2]);
                else
                    System.out.println(-1);
            }
        }
    }

}