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<long long> v;
long long GCD(long long a, long long b) {
  if (b == 0)
    return a;
  else
    return GCD(b, a % b);
}
void printDivisors(long long n) {
  for (long long i = 1; i <= sqrt(n); i++) {
    if (n % i == 0) {
      if (n / i == i)
        v.push_back(i);
      else {
        v.push_back(i);
        v.push_back(n / i);
      }
    }
  }
}
int main() {
  long long a, b;
  cin >> a >> b;
  long long t;
  cin >> t;
  long long g = GCD(a, b);
  printDivisors(g);
  sort(v.begin(), v.end());
  for (int i = 0; i < t; i++) {
    long long l, u;
    cin >> l >> u;
    vector<long long>::iterator it = lower_bound(v.begin(), v.end(), u);
    if (it == v.end() || *it > u) it--;
    if (*it >= l)
      cout << *it << '\n';
    else
      cout << -1 << '\n';
  }
  return 0;
}