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>
const double PI = 3.14;
using namespace std;
void R() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
const double EPS = 1e-6;
const int N = 1e7 + 3;
int DCNP(double a, double b) {
  if (fabs(a - b) < EPS) return 0;
  return (a < b ? -1 : 1);
}
int gcd(int x, int y) { return (!y) ? x : gcd(y, x % y); }
int main() {
  R();
  int a, b, q, m, n;
  cin >> a >> b >> q;
  vector<int> ans;
  n = gcd(a, b);
  for (int i = 1; i <= sqrt(n); i++) {
    if (n % i == 0) {
      ans.push_back(i);
      m = n / i;
      if (i != m && n % m == 0) ans.push_back(m);
    }
  }
  sort(ans.begin(), ans.end());
  while (q--) {
    int l, r;
    cin >> l >> r;
    auto it = upper_bound(ans.begin(), ans.end(), r) - ans.begin();
    --it;
    if (it >= 0 && it < ans.size() && ans[it] >= l && ans[it] <= r)
      cout << ans[it] << "\n";
    else
      cout << "-1\n";
  }
}