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() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int a, b, n;
  cin >> a >> b >> n;
  set<long long int> s, r, ans;
  for (long long int i = 1; i <= (sqrt(a)) + 1; ++i)
    if (a % i == 0) {
      s.insert(i);
      s.insert(a / i);
    }
  for (long long int i = 1; i <= (sqrt(b)) + 1; ++i)
    if (b % i == 0) {
      r.insert(i);
      r.insert(b / i);
    }
  set<long long int>::iterator it = s.begin();
  while (it != s.end()) {
    if (r.find(*it) != r.end()) ans.insert(*it);
    it++;
  }
  for (long long int i = 0; i <= n - 1; ++i) {
    long long int x, y;
    cin >> x >> y;
    if (ans.lower_bound(x) == ans.end())
      cout << -1 << endl;
    else if (ans.upper_bound(y) == ans.begin())
      cout << -1 << endl;
    else {
      it = ans.upper_bound(y);
      it--;
      if (*(it) >= x)
        cout << (*it) << endl;
      else
        cout << -1 << endl;
    }
  }
  return 0;
}