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;
set<int> S;
int a, b;
int q[7000000];
int cnt;
int l, r;
int t;
int main() {
  int i, j;
  cnt = 0;
  S.clear();
  cin >> a >> b;
  for (i = 1; i < 100000; i++) {
    if (a % i == 0) {
      if (b % i == 0) {
        S.insert(i);
      }
      j = a / i;
      if (b % j == 0) {
        S.insert(j);
      }
    }
  }
  set<int>::iterator it;
  for (it = S.begin(); it != S.end(); it++) q[cnt++] = *it;
  cin >> t;
  while (t--) {
    cin >> l >> r;
    int low = 0, high = cnt - 1;
    while (low + 1 < high) {
      int mid = (low + high) / 2;
      if (q[mid] <= r)
        low = mid;
      else
        high = mid - 1;
    }
    if (q[high] >= l && q[high] <= r)
      cout << q[high] << endl;
    else if (q[low] >= l && q[low] <= r)
      cout << q[low] << endl;
    else
      cout << -1 << endl;
  }
  return 0;
}