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 a, b, n, x, y, cnt, p[100009];
long long gcd(long long x, long long y) { return y ? gcd(y, x % y) : x; }
int main() {
  cin >> a >> b >> n, a = gcd(max(a, b), min(a, b));
  for (long long i = 1; i <= sqrt(a); i += 1)
    if (!(a % i)) p[cnt++] = i, p[cnt++] = a / i;
  sort(p, p + cnt);
  for (long long i = 0; i < n; i += 1) {
    cin >> x >> y;
    long long qwq = p[upper_bound(p, p + cnt, y) - p - 1];
    cout << (qwq >= x ? qwq : -1) << endl;
  }
  return 0;
}