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 n;
set<int> Hash;
int gcd(int x, int y) { return y ? gcd(y, x % y) : x; }
int main() {
  int a, b;
  scanf("%d%d", &a, &b);
  int g = gcd(a, b);
  scanf("%d", &n);
  Hash.clear();
  for (int i = 1; i * i <= g; i++)
    if (g % i == 0) {
      Hash.insert(i);
      Hash.insert(g / i);
    }
  for (int i = 0; i < n; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    int ret = -1;
    for (set<int>::iterator it = Hash.begin(); it != Hash.end(); it++)
      if (x <= *it && *it <= y && *it > ret) ret = *it;
    printf("%d\n", ret);
  }
}