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;
vector<long long> divisors;
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
void common_divisor(long long x, long long y) {
  long long z, m;
  for (long long i = 0; i < sqrt(max(x, y)); ++i) {
    z = 0, m = 0;
    if (!(x % (i + 1)) && !(y % (i + 1))) divisors.push_back(i + 1);
    if (!(x % (i + 1))) z = x / (i + 1);
    if (!(y % (i + 1))) m = y / (i + 1);
    if (m && !(x % m)) divisors.push_back(m);
    if (z && !(y % z)) divisors.push_back(z);
  }
}
int main() {
  long long x, y, s;
  cin >> x >> y;
  common_divisor(x, y);
  sort(divisors.begin(), divisors.end(), greater<long long>());
  long long t;
  cin >> t;
  while (t--) {
    cin >> x >> y;
    long long idx =
        lower_bound(divisors.begin(), divisors.end(), y, greater<long long>()) -
        divisors.begin();
    if (idx == -1 || divisors[idx] < x)
      printf("-1\n");
    else
      printf("%d\n", divisors[idx]);
  }
}