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 main() {
  int a, b;
  cin >> a >> b;
  vector<int> v;
  int sq = sqrt(min(a, b));
  for (int i = 1; i <= sq; i++) {
    if (a % i == 0 && b % i == 0) {
      v.push_back(i);
    }
    int x = a / i, y = b / i;
    if (a % x == 0 && b % x == 0) v.push_back(x);
    if (a % y == 0 && b % y == 0) v.push_back(y);
  }
  sort(v.begin(), v.end());
  int sz = v.size();
  int q;
  cin >> q;
  int lo, hi;
  while (q--) {
    cin >> lo >> hi;
    int ind = upper_bound(v.begin(), v.end(), hi) - v.begin();
    if (v[ind - 1] >= lo)
      cout << v[ind - 1] << endl;
    else
      cout << -1 << endl;
  }
}