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>
using namespace std;
int i, j, k, l, x, y, z, m, n, a, b, ans, p, q, r;
int numbers[800000];
int gcd(int x, int y) {
  if (y == 0)
    return x;
  else
    return gcd(y, x % y);
}
int main() {
  scanf("%d %d %d", &a, &b, &n);
  ans = gcd(a, b);
  m = sqrt(ans);
  for (i = 1, k = 0; i <= m; i++) {
    if (ans % i == 0) {
      p = ans / i;
      numbers[k] = p;
      k++;
      numbers[k] = i;
      k++;
    }
  }
  sort(numbers, numbers + k);
  p = 0;
  q = 0;
  for (i = 1; i <= n; i++) {
    scanf("%d %d", &x, &y);
    if (x > ans) {
      printf("-1\n");
      continue;
    }
    p = lower_bound(numbers, numbers + k, (x)) - numbers;
    q = lower_bound(numbers, numbers + k, (y)) - numbers;
    r = 0;
    for (j = q + 1; j >= p - 1; j--) {
      if (numbers[j] >= x && numbers[j] <= y) {
        printf("%d\n", numbers[j]);
        r = 1;
        break;
      }
    }
    if (!r) printf("-1\n");
  }
  return 0;
}