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 a, b, x, y, in, t, num[10006];
int gcd(int a, int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
int main() {
  cin >> a >> b;
  in = gcd(a, b);
  for (int i = 1; i <= sqrt(in); i++) {
    if (in % i) continue;
    num[++num[0]] = i;
    if (i * i != in) num[++num[0]] = in / i;
  }
  sort(num + 1, num + 1 + num[0]);
  cin >> t;
  while (t--) {
    cin >> x >> y;
    int l = 1, r = num[0], ans = -1, mid;
    while (l <= r) {
      mid = (l + r) / 2;
      if (num[mid] <= y) {
        l = mid + 1, ans = num[mid];
      } else
        r = mid - 1;
    }
    if (ans < x)
      cout << "-1" << endl;
    else
      cout << ans << endl;
  }
  return 0;
}