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;
map<int, int> mp;
int gcd(int a, int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
void fillmap(int gc) {
  for (int i = 1; i * i <= gc; i++) {
    if (gc % i == 0) mp[i] = gc / i;
  }
}
int main() {
  int a, b, n, gc;
  cin >> a >> b;
  cin >> n;
  gc = gcd(a, b);
  fillmap(gc);
  for (int i = 0; i < n; i++) {
    int l, r;
    cin >> l >> r;
    int max = -1;
    for (std::map<int, int>::iterator it = mp.begin(); it != mp.end(); ++it) {
      if (it->first >= l && it->first <= r && it->first > max) max = it->first;
      if (it->second >= l && it->second <= r && it->second > max)
        max = it->second;
    }
    cout << max << endl;
  }
  return 0;
}