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;
const int mod = 1e9 + 7;
int main(int argc, char const *argv[]) {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long a, b;
  cin >> a >> b;
  std::map<long long, int> h;
  for (long long i = 1; i * i <= a; ++i) {
    if (a % i == 0) {
      h[i] += 1;
      if (a / i != i) h[a / i] += 1;
    }
  }
  for (long long i = 1; i * i <= b; ++i) {
    if (b % i == 0) {
      h[i] += 1;
      if (b / i != i) h[b / i] += 1;
    }
  }
  std::vector<long long> v;
  for (std::map<long long, int>::iterator it = h.begin(); it != h.end(); it++) {
    if (it->second == 2) v.push_back(it->first);
  }
  sort(v.rbegin(), v.rend());
  int q;
  cin >> q;
  while (q--) {
    long long m, n;
    cin >> m >> n;
    bool b = 0;
    for (int i = 0; i < v.size(); ++i) {
      if (v[i] >= m && v[i] <= n) {
        cout << v[i] << endl;
        b = 1;
        break;
      }
    }
    if (!b) cout << -1 << endl;
  }
  return 0;
}