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>
long as, x, y, p, d, m, n, i, l;
int main() {
  scanf("%ld%ld", &x, &y);
  do {
    if (x > y)
      x = x % y;
    else
      y = y % x;
  } while ((x != 0) && (y != 0));
  scanf("%ld", &n);
  p = x + y;
  for (i = 1; i <= n; i++) {
    scanf("%ld%ld", &x, &y);
    if (x > p)
      printf("-1\n");
    else {
      l = p / y;
      if (l == 0) l = 1;
      l--;
      as = 0;
      while (x > -1) {
        l++;
        d = p % l;
        m = p / l;
        if (d == 0) {
          if ((x <= m) && (m <= y)) {
            as = 1;
            printf("%ld\n", m);
            break;
          }
        }
        if (m < x) break;
      }
      if (as == 0) printf("-1\n");
    }
  }
}