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;
long long GCD(long long a, long long b) { return (a) ? GCD(b % a, a) : b; }
long long LCM(long long a, long long b) { return a * b / GCD(a, b); }
double power(double a, long long b) {
  double x = 1;
  for (int i = 0; i < b; i++) {
    x = x * a;
  }
  return x;
}
int main() {
  long long a, b, n, low, high;
  cin >> a >> b >> n;
  vector<long long> fact;
  long long num = GCD(a, b);
  long long i;
  for (i = 1; i * i < num; i++) {
    if (num % i == 0) {
      fact.push_back(i);
      fact.push_back(num / i);
    }
  }
  if (i * i == num) fact.push_back(i);
  long long S = fact.size();
  sort(fact.begin(), fact.end());
  while (n--) {
    cin >> low >> high;
    long long l = 0;
    long long r = S - 1;
    long long ans = -1;
    while (l <= r) {
      long long mid = (l + r) / 2;
      if (fact[mid] >= low && fact[mid] <= high) {
        ans = fact[mid];
        l = mid + 1;
      } else if (fact[mid] < low) {
        l = mid + 1;
      } else {
        r = mid - 1;
      }
    }
    cout << ans << endl;
  }
  return 0;
}