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 i, j, k;
  int n;
  long long int a, b, x, m;
  long long int l, r, m2;
  scanf("%I64d%I64d", &a, &b);
  scanf("%d", &n);
  m = min(a, b);
  m2 = max(a, b);
  set<long long int> s;
  for (i = 1; i <= sqrt(m); i++) {
    if (m % i == 0) s.insert(m / i), s.insert(i);
  }
  vector<long long int> v;
  set<long long int>::iterator p = s.begin();
  while (p != s.end()) {
    v.push_back(*p);
    p++;
  }
  for (i = 0; i < n; i++) {
    scanf("%I64d%I64d", &l, &r);
    x = -1;
    for (j = 0; j < v.size(); j++) {
      if (v[j] > r) break;
      if (v[j] >= l && v[j] <= r && m2 % v[j] == 0) x = v[j];
    }
    printf("%I64d\n", x);
  }
  return 0;
}