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(void) {
  int a, b;
  cin >> a >> b;
  if (a < b) swap(a, b);
  vector<int> v;
  v.push_back(1);
  vector<int>::iterator it1, it2;
  for (int i = 1; i * i <= a; i++) {
    if (a % i == 0) {
      if (b % i == 0) v.push_back(i);
      if (b % (a / i) == 0 && i * i != a) v.push_back(a / i);
    }
  }
  sort(v.begin(), v.end());
  int t, l, r;
  cin >> t;
  while (t--) {
    cin >> l >> r;
    it1 = lower_bound(v.begin(), v.end(), l);
    if ((*it1) <= r && it1 != v.end()) {
      it2 = upper_bound(v.begin(), v.end(), r);
      cout << *(it2 - 1) << endl;
    } else
      cout << "-1" << endl;
  }
  return 0;
}