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 pgcd(int a, int b) {
  if (b % a == 0) return a;
  pgcd(b % a, a);
}
int main() {
  int o = 0, gcd, az[1001], t[5000], j, k = 0, m, x, y, i, a, b, n;
  cin >> a >> b >> n;
  gcd = pgcd(a, b);
  for (i = 1; i * i <= gcd; i++) {
    if (gcd % i == 0) {
      t[k] = i;
      k++;
      t[k] = gcd / i;
      k++;
    }
  }
  for (i = 0; i < n; i++) {
    m = -1;
    cin >> x >> y;
    for (j = 0; j < k; j++) {
      if (t[j] <= y && t[j] >= x) m = max(m, t[j]);
    }
    cout << m << endl;
  }
}