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;
vector<int> g;
int gcd(int a, int b) { return !b ? a : gcd(b, a % b); }
int main(void) {
  int a, b;
  cin >> a >> b;
  int gc = gcd(a, b);
  for (int i = 1; i * i <= gc; ++i)
    if (!(gc % i)) {
      g.push_back(i);
      if (i * i != gc) g.push_back(gc / i);
    }
  sort(g.begin(), g.end());
  int n;
  cin >> n;
  while (n--) {
    cin >> a >> b;
    int ans = -1;
    for (vector<int>::iterator it = g.begin(); it != g.end(); ++it)
      if (a <= *it && *it <= b) ans = max(ans, *it);
    cout << ans << '\n';
  }
  return 0;
}