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 gcd(int a, int b) {
  if (!b) return a;
  return gcd(b, a % b);
}
int lcm(int a, int b) { return a / gcd(a, b) * b; }
bool isprime(long long n) {
  if (n == 0 || n == 1) return 0;
  for (long long i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      return 0;
    }
  }
  return 1;
}
int main() {
  int a, b, n;
  cin >> a >> b;
  set<int> s1, s2, s;
  for (int i = 1; i * i <= a; i++)
    if (a % i == 0) s1.insert(i), s1.insert(a / i);
  for (int i = 1; i * i <= b; i++)
    if (b % i == 0) s2.insert(i), s2.insert(b / i);
  set_intersection(s1.begin(), s1.end(), s2.begin(), s2.end(),
                   inserter(s, s.begin()));
  cin >> n;
  while (n--) {
    int k = -1, l, r;
    cin >> l >> r;
    for (auto it : s) {
      if (it <= r) {
        if (it >= l) k = it;
      } else
        break;
    }
    cout << k << '\n';
  }
  return 0;
}