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;
long long int ceil2(long long int a, long long int b) {
  return (a + b - 1) / b;
}
long long int hcf(long long int n1, long long int n2) {
  if (n2 != 0)
    return hcf(n2, n1 % n2);
  else
    return n1;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int q;
  long long int a, b, li, hi;
  cin >> a >> b;
  cin >> q;
  int gcd = hcf(a, b);
  map<int, int> m;
  vector<long long int> ice;
  while (gcd % 2 == 0) {
    ice.push_back(2);
    gcd /= 2;
  }
  for (int i = 3; i * i <= gcd; i++) {
    while (gcd % i == 0) {
      ice.push_back(i);
      gcd /= i;
    }
  }
  if (gcd > 1) {
    ice.push_back(gcd);
  }
  int sz = ice.size();
  vector<long long int> requim;
  requim.push_back(1);
  int zx = requim.size();
  for (int i = 0; i < sz; i++) {
    for (int j = 0; j < zx; j++) {
      requim.push_back(requim[j] * ice[i]);
    }
    zx = requim.size();
  }
  sort(requim.begin(), requim.end());
  for (int i = 0; i < q; i++) {
    cin >> li >> hi;
    int pos = upper_bound(requim.begin(), requim.end(), hi) - requim.begin();
    if (pos == (zx + 1)) {
      cout << -1 << endl;
    } else if (pos == 0) {
      cout << -1 << endl;
    } else if (requim[pos - 1] >= li) {
      cout << requim[pos - 1] << endl;
    } else {
      cout << -1 << endl;
    }
  }
  return 0;
}