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> divisors;
int gcd(int m, int n) {
  while (n != 0) {
    int t = m % n;
    m = n;
    n = t;
  }
  return m;
}
void get_divisors(int g) {
  int rt = sqrt(g);
  for (int i = 1; i <= rt; ++i) {
    if (g % i == 0) {
      divisors.push_back(i);
      divisors.push_back(g / i);
    }
  }
  if (divisors.back() == rt) divisors.pop_back();
  sort(divisors.begin(), divisors.end());
}
int main() {
  int a, b;
  scanf("%d %d", &a, &b);
  get_divisors(gcd(a, b));
  int n;
  scanf("%d", &n);
  int low, high;
  while (n--) {
    scanf("%d %d", &low, &high);
    int idx =
        upper_bound(divisors.begin(), divisors.end(), high) - divisors.begin();
    printf("%d\n",
           ((idx > 0 && low <= divisors[idx - 1]) ? divisors[idx - 1] : -1));
  }
  return 0;
}