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() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  int a, b;
  cin >> a >> b;
  vector<int> v;
  vector<int> div;
  map<int, int> m;
  map<int, int> done;
  for (int i = 1; i * i <= a; i++) {
    if (a % i == 0) {
      div.push_back(i);
      m[i]++;
      if (i != a / i) {
        div.push_back(a / i);
        m[a / i]++;
      }
    }
  }
  for (int i = 1; i * i <= b; i++) {
    if (b % i == 0) {
      div.push_back(i);
      m[i]++;
      if (i != b / i) {
        div.push_back(b / i);
        m[b / i]++;
      }
    }
  }
  sort(div.begin(), div.end());
  v.push_back(div[0]);
  done[div[0]]++;
  for (int i = 1; i < div.size(); i++) {
    if (done[div[i]] == 0 && m[div[i]] >= 2) {
      v.push_back(div[i]);
      done[div[i]]++;
    }
  }
  int q;
  cin >> q;
  while (q--) {
    int l, r;
    cin >> l >> r;
    bool found = false;
    for (int i = v.size() - 1; i >= 0; i--) {
      if (v[i] >= l && v[i] <= r) {
        cout << v[i] << "\n";
        found = true;
        break;
      }
    }
    if (!found) cout << "-1\n";
  }
  return 0;
}