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;
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
int main() {
  long long a, b;
  cin >> a >> b;
  long long r = gcd(a, b);
  long long n;
  cin >> n;
  while (n--) {
    long long x, y;
    cin >> x >> y;
    long long q = (r / y) + 1;
    if (x > r)
      cout << -1 << endl;
    else if (r % x != 0 && r % y != 0 && r / x == r / y)
      cout << -1 << endl;
    else if (r % y == 0)
      cout << y << endl;
    else {
      for (long long j = q; j <= (r / x); j++) {
        if (r % j == 0) {
          cout << r / j << endl;
          break;
        } else if (j == (r / x)) {
          cout << -1 << endl;
          break;
        }
      }
    }
  }
  return 0;
}