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 main() {
  unsigned long a, b, h, l, c, d;
  int n;
  cin >> a >> b >> n;
  unsigned long t;
  while (b != 0) {
    t = b;
    b = a % b;
    a = t;
  }
  while (n--) {
    cin >> l >> h;
    if (t <= h) {
      if (t < l) {
        cout << "-1" << endl;
      } else {
        cout << t << endl;
      }
    } else {
      if (t % h == 0) {
        c = t / h;
      } else {
        c = t / h + 1;
      }
      d = t / l;
      while (t % c != 0 && c <= d) {
        c++;
      }
      if (c <= d) {
        cout << t / c << "\n";
      } else
        cout << "-1" << endl;
    }
  }
}