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 MOD = 1e9 + 7;
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
void solve() {
  long long a, b;
  cin >> a >> b;
  long long g = gcd(a, b);
  vector<long long> fc;
  fc.push_back(1);
  fc.push_back(g);
  for (long long i = 2; i * i <= g; i++) {
    if (g % i == 0) {
      fc.push_back(i);
      fc.push_back(g / i);
    }
  }
  sort(fc.begin(), fc.end());
  long long n;
  cin >> n;
  while (n--) {
    long long l, r;
    cin >> l >> r;
    long long ans = -1;
    long long low = 0, high = (long long)fc.size() - 1;
    while (high >= low) {
      long long mid = low + (high - low) / 2;
      if (fc[mid] < l)
        low = mid + 1;
      else if (fc[mid] > r)
        high = mid - 1;
      else {
        ans = fc[mid];
        low = mid + 1;
      }
    }
    cout << ans << "\n";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  long long o = 1;
  while (t--) {
    solve();
    o++;
  }
}