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() {
  ios_base::sync_with_stdio(false);
  int a, b, n;
  vector<int> div;
  cin >> a >> b >> n;
  if (b > a) swap(a, b);
  for (int d = 1; d * d <= b; d++) {
    if (a % d == 0 and b % d == 0) div.push_back(d);
    int d2 = -1;
    if (b % d == 0 and b / d != d) d2 = b / d;
    if (d2 != -1)
      if (a % d2 == 0) div.push_back(d2);
  }
  sort(div.begin(), div.end());
  while (n--) {
    int l, r;
    cin >> l >> r;
    auto it = upper_bound(div.begin(), div.end(), r);
    it--;
    if (l <= *it and *it <= r)
      cout << *it << endl;
    else
      cout << "-1\n";
  }
}