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 CPP):

Solution:
#include <bits/stdc++.h>
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
int main() {
  int a, b, g, i, n, max;
  scanf("%d%d%d", &a, &b, &n);
  g = gcd(a, b);
  while (n-- > 0) {
    int low, high;
    scanf("%d%d", &low, &high);
    max = -1;
    for (i = 1; i * i <= g; i++) {
      if (g % i == 0 && i >= low && i <= high && max < g / i) max = i;
      if (g % i == 0 && g / i >= low && g / i <= high && max < g / i)
        max = g / i;
    }
    printf("%d\n", max);
  }
  return 0;
}