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 N = 1e5 + 9;
long long a, b, n, dv[N], l, r, sz;
long long bs(long long l, long long r) {
  long long bas = 0;
  long long son = sz + 1;
  while (bas + 1 < son) {
    long long orta = bas + son >> 1;
    if (dv[orta] < l)
      bas = orta;
    else if (dv[orta] > r)
      son = orta;
    else if (dv[orta] <= r and dv[orta] >= l)
      bas = orta;
  }
  return (dv[bas] < l or dv[bas] > r) ? -1 : dv[bas];
}
int main() {
  cin >> a >> b >> n;
  for (long long i = 1; i * i <= max(a, b); i++) {
    if (a % i == 0 and b % i == 0) {
      dv[++sz] = i;
      if (b % (a / i) == 0 and i * i != max(a, b)) dv[++sz] = a / i;
      if (a % (b / i) == 0) dv[++sz] = b / i;
    } else if (a % (b / i) == 0 and b % i == 0)
      dv[++sz] = b / i;
    else if (b % (a / i) == 0 and a % i == 0)
      dv[++sz] = a / i;
  }
  sort(dv + 1, dv + sz + 1);
  for (long long i = 1; i <= n; i++) {
    cin >> l >> r;
    cout << bs(l, r) << endl;
  }
  return 0;
}