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.Scanner;
import java.util.StringTokenizer;

public class Main {

    public static void main(String[] args) throws IOException {
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st;
        st = new StringTokenizer(in.readLine());
        int a = Integer.parseInt(st.nextToken());
        int b = Integer.parseInt(st.nextToken());
        int n = Integer.parseInt(in.readLine());
        int min = Math.min(a, b) + 10;
        int s = (int) Math.sqrt(min);
        int[] gcds = new int[4 * s];
        int r = 0;
        for (int i = 1; i <= s; i++) {
            if (a % i == 0 && b % (a / i) == 0)
                gcds[r++] = a / i;
            if (b % i == 0 && a % (b / i) == 0)
                gcds[r++] = b / i;
            if (a % i == 0 && b % i == 0)
                gcds[r++] = i;
        }
        int f, t;
        for (int i = 0; i < n; i++) {
            st = new StringTokenizer(in.readLine());
            f = Integer.parseInt(st.nextToken());
            t = Integer.parseInt(st.nextToken());
            int max = -1;
            for (int j = 0; j < r; j++) {
                if (gcds[j] >= f && gcds[j] <= t)
                    max = Math.max(max, gcds[j]);
            }
            System.out.println(max);
        }
    }
}