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<int> x;
void divs(int a, int b) {
  if (a > b) swap(a, b);
  long long i = 1;
  for (; i * i < a; ++i)
    if (a % i == 0) {
      if (b % i == 0) x.emplace_back(i);
      if (b % (a / i) == 0) x.emplace_back(a / i);
    }
  if (i * i == a && b % i == 0) x.emplace_back(i);
  sort(x.begin(), x.end());
}
int F(int f, int l) {
  int p = lower_bound(x.begin(), x.end(), l) - begin(x);
  if (p == ((int)((x).size())) || x[p] > l) --p;
  return (x[p] >= f ? x[p] : -1);
}
int main() {
  int a, b;
  scanf("%d %d", &a, &b);
  divs(a, b);
  int q;
  scanf("%d", &q);
  while (q--) {
    int f, l;
    scanf("%d %d", &f, &l);
    printf("%d\n", F(f, l));
  }
  return 0;
}