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() {
  vector<long long int> answer;
  long long int a, b, n, j, low, high, k;
  cin >> a >> b;
  if (a > b) {
    swap(a, b);
  }
  cin >> n;
  for (j = 1; j <= int(sqrt(a)) + 1; j++) {
    if (a % j == 0) {
      if (b % j == 0) {
        answer.push_back(j);
      }
      if (b % (a / j) == 0) {
        answer.push_back(a / j);
      }
    }
  }
  sort(answer.begin(), answer.end());
  for (j = 1; j <= n; j++) {
    cin >> low >> high;
    k = (upper_bound(answer.begin(), answer.end(), high) - answer.begin());
    if (answer[k - 1] >= low) {
      cout << answer[k - 1] << endl;
    } else {
      cout << "-1" << endl;
    }
  }
}