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 main() {
  int n, x, y, l, h, i, max;
  scanf("%d %d %d", &x, &y, &n);
  while (x != y) {
    if (x > y)
      x -= y;
    else
      y -= x;
  }
  while (n--) {
    scanf("%d %d", &l, &h);
    for (i = 1, max = -1; i * i <= x; i++) {
      if (x % i == 0 && i >= l && i <= h && x / i > max) max = i;
      if (x % i == 0 && x / i >= l && x / i <= h && x / i > max) max = x / i;
    }
    printf("%d\n", max);
  }
  return 0;
}