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;
int main() {
  long long int a, b;
  cin >> a >> b;
  vector<long long int> v;
  map<long long int, long long int> mp;
  for (int i = 1; i * i <= a; i++) {
    if (a % i == 0) {
      mp[i]++;
      if (i != a / i) {
        mp[a / i]++;
      }
    }
  }
  for (int i = 1; i * i <= b; i++) {
    if (b % i == 0) {
      mp[i]++;
      if (i != b / i) {
        mp[b / i]++;
      }
    }
  }
  for (auto x : mp) {
    if (x.second == 2) {
      v.push_back(x.first);
    }
  }
  int q;
  cin >> q;
  while (q--) {
    long long int l, r;
    cin >> l >> r;
    int id = lower_bound(v.begin(), v.end(), l) - v.begin();
    int id2 = upper_bound(v.begin(), v.end(), r) - v.begin();
    id2--;
    if (id > id2) {
      cout << "-1\n";
    } else {
      cout << v[id2] << "\n";
    }
  }
  return 0;
}