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> graph[200100];
int vis[200100];
long long gcd(long long a, long long b) {
  if (a > b) gcd(b, a);
  if (a == 0) return b;
  return (gcd(b % a, a));
}
int main() {
  long long a, b;
  cin >> a >> b;
  long long g = gcd(a, b);
  int n;
  cin >> n;
  vector<long long> factors;
  for (long long i = 1; i <= sqrt(g); i++) {
    if (g % i == 0) {
      factors.push_back(i);
      if (i != g / i) factors.push_back(g / i);
    }
  }
  sort(factors.begin(), factors.end());
  for (int i = 0; i < n; i++) {
    long long x, y;
    cin >> x >> y;
    if (x > g)
      cout << -1 << endl;
    else if (x <= g && y >= g)
      cout << g << endl;
    else {
      int start = 0;
      int end = factors.size() - 1;
      int mid = 0;
      long long ans = -1;
      while (start <= end) {
        mid = (start + end) / 2;
        if (factors[mid] <= y) {
          start = mid + 1;
          if (factors[mid] >= x) ans = factors[mid];
        } else
          end = mid - 1;
      }
      cout << ans << endl;
    }
  }
  return 0;
}