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 gcd(int a, int b) {
  while (b) {
    auto tmp = a;
    a = b;
    b = tmp % b;
  }
  return a;
}
void oneCase() {
  int a, b;
  cin >> a >> b;
  int n;
  cin >> n;
  auto d = gcd(a, b);
  set<int> div;
  for (int i = 1; i * i <= d; ++i) {
    if (d % i == 0) {
      div.insert(i);
      div.insert(d / i);
    }
  }
  while (n--) {
    int l, h;
    cin >> l >> h;
    auto p = div.upper_bound(h);
    if (p == div.begin()) {
      cout << "-1\n";
      continue;
    }
    --p;
    if (*p >= l) {
      cout << *p << "\n";
    } else {
      cout << "-1\n";
    }
  }
}
int main() {
  int t = 1;
  while (t--) {
    oneCase();
  }
  return 0;
}