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 pwr(long long base, long long p, long long mod = 1000000000) {
  long long ans = 1;
  while (p) {
    if (p & 1) ans = (ans * base) % 1000000000;
    base = (base * base) % 1000000000;
    p /= 2;
  }
  return ans;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
bool isPrime[10000005];
void sieve(int n) {
  for (int i = 0; i <= n; ++i) {
    isPrime[i] = true;
  }
  isPrime[0] = false;
  isPrime[1] = false;
  for (int i = 2; i * i <= n; ++i) {
    if (isPrime[i] == true) {
      for (int j = i * i; j <= n; j += i) isPrime[j] = false;
    }
  }
}
bool cmp(const pair<int, int> &a, const pair<int, int> &b) {
  return (a.first > b.first);
}
const long long N = 2e5 + 5;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long a, b, l, h, q;
  cin >> a >> b >> q;
  set<long long> s;
  set<long long>::iterator it, itr;
  long long g = gcd(a, b);
  for (long long i = 1; i * i <= g; i++) {
    if (g % i == 0) {
      s.insert(i);
      s.insert(g / i);
    }
  }
  while (q--) {
    long long x, y;
    cin >> x >> y;
    it = s.lower_bound(x);
    if (it == s.end())
      cout << -1 << endl;
    else {
      itr = s.upper_bound(y);
      itr--;
      long long av = *it;
      long long bv = *itr;
      if (av > bv)
        cout << -1 << endl;
      else
        cout << bv << endl;
    }
  }
  return 0;
}