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;
long long int apSum(long long int a, long long int n, long long int d = 1) {
  long long int res = (2 * a + (n - 1) * d) * n;
  res /= 2;
  return res;
}
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
bool isPrime(long long int n) {
  for (long long int i = 2; i * i <= n; i++) {
    if (n % i == 0) return false;
  }
  return true;
}
void solve() {
  long long int a, b, n;
  cin >> a >> b >> n;
  long long int limit = min(a, b);
  b = max(a, b);
  a = limit;
  vector<long long int> vec;
  for (long long int i = 1; i * i <= limit; i++) {
    if (a % i == 0) {
      if (b % i == 0) vec.push_back(i);
      long long int v = a / i;
      if (b % v == 0) vec.push_back(v);
    }
  }
  sort(vec.begin(), vec.end());
  while (n--) {
    long long int x, y;
    cin >> x >> y;
    long long int ans = -1;
    long long int i = 0;
    for (i; i < vec.size(); i++) {
      if (vec[i] >= x && vec[i] <= y) {
        ans = max(ans, vec[i]);
      }
      if (vec[i] > y) break;
    }
    cout << ans << endl;
  }
}
int32_t main() {
  long long int t = 1;
  while (t--) {
    solve();
  }
}