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 (a == 0) return b;
  return gcd(b % a, a);
}
vector<long long int> commDiv(long long int a, long long int b) {
  int n = gcd(a, b);
  vector<long long int> ans;
  for (int i = 1; i <= sqrt(n); i++) {
    if (n % i == 0) {
      ans.push_back(n / i);
      ans.push_back(i);
    }
  }
  sort(ans.begin(), ans.end());
  ans.erase(unique(ans.begin(), ans.end()), ans.end());
  return ans;
}
int main() {
  long long int a, b;
  cin >> a >> b;
  vector<long long int> nums = commDiv(a, b);
  int n;
  cin >> n;
  long long int x, y;
  for (int i = 0; i < n; i++) {
    cin >> x >> y;
    auto it = lower_bound(nums.begin(), nums.end(), y);
    if (*it != y) it--;
    if (*it < x)
      cout << -1 << endl;
    else
      cout << *it << endl;
  }
}