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;
template <class T>
T _abs(T n) {
  return (n < 0 ? -n : n);
}
template <class T>
T _max(T a, T b) {
  return (!(a < b) ? a : b);
}
template <class T>
T _min(T a, T b) {
  return (a < b ? a : b);
}
template <class T>
T sq(T x) {
  return x * x;
}
template <class T>
T gcd(T a, T b) {
  return (b != 0 ? gcd<T>(b, a % b) : a);
}
template <class T>
T lcm(T a, T b) {
  return (a / gcd<T>(a, b) * b);
}
template <class T>
bool inside(T a, T b, T c) {
  return a <= b && b <= c;
}
const int MAXN = 200005;
const int maxnum = 100010;
const int INF = 0x3f3f3f3f;
const double eps = 1e-9;
const int mod = 1e9;
using namespace std;
set<int> s;
set<int>::iterator it;
int a, b, n;
int main() {
  cin >> a >> b;
  cin >> n;
  int gg = gcd(a, b);
  for (int i = 1; i * i <= gg; i++) {
    if (gg % i == 0) {
      s.insert(i);
      s.insert(gg / i);
    }
  }
  while (n--) {
    cin >> a >> b;
    it = s.upper_bound(b);
    if (*--it < a)
      printf("-1\n");
    else
      printf("%d\n", *it);
  }
  return 0;
}