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;
unordered_map<long long, int> mp;
vector<long long> divis;
void divi(long long num) {
  for (long long i = 1; i * i <= num; ++i) {
    if (num % i == 0LL) {
      mp[i]++;
      if (mp[i] == 2) divis.emplace_back(i);
      if (num / i != i) {
        mp[num / i]++;
        if (mp[num / i] == 2) divis.emplace_back(num / i);
      }
    }
  }
}
int main() {
  long long a, b, lo, hi;
  scanf("%lld%lld", &a, &b);
  divi(a);
  divi(b);
  sort(divis.begin(), divis.end());
  int n;
  scanf("%d", &n);
  while (n--) {
    scanf("%lld%lld", &lo, &hi);
    int id = upper_bound(divis.begin(), divis.end(), hi) - divis.begin();
    id--;
    if (id < 0 || divis[id] < lo || divis[id] > hi)
      puts("-1");
    else
      printf("%I64d\n", divis[id]);
  }
  return 0;
}