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;
pair<int, int> p;
vector<int> v;
void GenDiv(int x, int y) {
  for (int i = 1; i * i <= x; i++) {
    if (x % i == 0) {
      if (y % i == 0) v.push_back(i);
      if (y % (x / i) == 0) v.push_back(x / i);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int n, a, b;
  cin >> a >> b >> n;
  GenDiv(a, b);
  sort(v.begin(), v.end());
  for (int i = 0; i < n; ++i) {
    cin >> p.first >> p.second;
    int idx = upper_bound(v.begin(), v.end(), p.second) - v.begin();
    cout << (v[idx - 1] >= p.first ? v[idx - 1] : -1) << endl;
  }
  return 0;
}