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;
void preprocess(int a, int b, set<int>& q, set<int>& q1) {
  for (long long i = 2; i * i <= a; ++i) {
    if (a % i == 0) {
      q.insert(i);
      q.insert(a / i);
    }
  }
  for (long long i = 2; i * i <= b; ++i) {
    if (b % i == 0) {
      q1.insert(i);
      q1.insert(b / i);
    }
  }
}
int main() {
  int a, b;
  cin >> a >> b;
  set<int> q, q1;
  preprocess(a, b, q, q1);
  int Q;
  cin >> Q;
  q.insert(1);
  q1.insert(1);
  q.insert(a);
  q1.insert(b);
  while (Q--) {
    int le, ri;
    cin >> le >> ri;
    auto it = q.lower_bound(le);
    auto it1 = q1.lower_bound(le);
    int ans = -1;
    while (it != q.end() && it1 != q1.end() && *it <= ri && *it1 <= ri) {
      int v = *it, u = *it1;
      if (v == u) {
        ans = max(v, ans);
        ++it;
        ++it1;
      } else if (v < u) {
        ++it;
      } else {
        ++it1;
      }
    }
    cout << ans << '\n';
  }
  return 0;
}