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() {
  long long a, b, n;
  cin >> a >> b >> n;
  map<long long, bool> A;
  map<long long, bool> B;
  for (long long i = 1; i <= sqrt(a); i++) {
    if (a % i == 0) {
      A[i] = 1;
      A[a / i] = 1;
    }
  }
  for (long long i = 1; i <= sqrt(b); i++) {
    if (b % i == 0) {
      B[i] = 1;
      B[b / i] = 1;
    }
  }
  long long M = min(a, b);
  if (M == a) {
    for (long long i = 0; i < n; i++) {
      long long R, L;
      long long d = -1;
      cin >> L >> R;
      for (auto j = A.begin(); j != A.end(); j++) {
        if (B[j->first] && j->first >= L && j->first <= R) {
          d = j->first;
        }
      }
      cout << d << endl;
    }
  } else {
    for (long long i = 0; i < n; i++) {
      long long R, L;
      long long d = -1;
      cin >> L >> R;
      for (auto j = B.begin(); j != B.end(); j++) {
        if (A[j->first] && j->first >= L && j->first <= R) {
          d = j->first;
        }
      }
      cout << d << endl;
    }
  }
  return 0;
}