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 A, B, N, Low, High;
set<int> Div;
int GCD(int A, int B) {
  if (!B) return A;
  return GCD(B, A % B);
}
int main() {
  scanf("%i %i", &A, &B);
  int D = GCD(A, B);
  for (int i = 1; 1LL * i * i <= D; ++i)
    if (D % i == 0) Div.insert(i), Div.insert(D / i);
  scanf("%i", &N);
  for (; N; N--) {
    scanf("%i %i", &Low, &High);
    set<int>::iterator it = Div.lower_bound(High);
    int Val = -1;
    if (it == Div.end() || !(Low <= *it && *it <= High)) {
      if (it != Div.begin()) --it;
      if (Low <= *it && *it <= High) Val = *it;
    } else if (Low <= *it && *it <= High)
      Val = *it;
    printf("%i\n", Val);
  }
}