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 a, b, q, x, y, t, cnt = 0, c[100099];
int gcd(int x, int y) {
  if (y == 0)
    return x;
  else
    return gcd(y, x % y);
}
int main() {
  cin >> a >> b >> q;
  t = gcd(a, b);
  for (int i = 1; i * i < t; i++) {
    if (t % i == 0) c[++cnt] = i, c[++cnt] = t / i;
  }
  if (int(sqrt(t)) * int(sqrt(t)) == t) c[++cnt] = int(sqrt(t));
  sort(c + 1, c + cnt + 1);
  while (q--) {
    cin >> x >> y;
    int L = 1, R = cnt, k;
    while (L <= R) {
      int Mid = (L + R) >> 1;
      if (c[Mid] <= y)
        k = Mid, L = Mid + 1;
      else
        R = Mid - 1;
    }
    if (c[k] >= x)
      cout << c[k] << endl;
    else
      cout << "-1" << endl;
  }
  return 0;
}