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> primes;
int gcd(int a, int b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
int find_lindex(int k) {
  int l = -1, r = primes.size();
  while (r - l > 1) {
    int mid = (l + r) / 2;
    if (primes[mid] >= k)
      r = mid;
    else
      l = mid;
  }
  return r;
}
int find_rindex(int k) {
  int l = -1, r = primes.size();
  while (r - l > 1) {
    int mid = (l + r) / 2;
    if (primes[mid] <= k)
      l = mid;
    else
      r = mid;
  }
  return l;
}
int main() {
  int a, b;
  cin >> a >> b;
  int c = gcd(a, b);
  for (int i = 1; i * i <= c; ++i) {
    if (c % i == 0) {
      primes.push_back(i);
      primes.push_back(c / i);
    }
  }
  sort(primes.begin(), primes.end());
  int n;
  cin >> n;
  for (int i = 0; i < n; ++i) {
    int l, r;
    cin >> l >> r;
    int l1 = find_lindex(l), r1 = find_rindex(r);
    if (r1 < l1)
      cout << "-1\n";
    else
      cout << primes[r1] << "\n";
  }
}