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 maxn = 6e5 + 1;
long long a, b;
int n;
long long gcd(long long a, long long b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
int main() {
  cin >> a >> b;
  ;
  long long val = gcd(a, b);
  vector<long long> gcdtbl;
  for (int i = 1; i * i <= val; i++) {
    if (val % i == 0) {
      gcdtbl.push_back(i);
      if (i * i != val) gcdtbl.push_back(val / i);
    }
  }
  sort(gcdtbl.begin(), gcdtbl.end());
  cin >> n;
  ;
  while (n--) {
    long long low, high;
    cin >> low >> high;
    ;
    int left = 0, right = gcdtbl.size() - 1;
    long long ans = -1;
    while (left <= right) {
      int mid = (left + right) / 2;
      if (gcdtbl[mid] >= low and gcdtbl[mid] <= high) {
        ans = gcdtbl[mid];
        left = mid + 1;
        continue;
      }
      if (gcdtbl[mid] <= low) {
        left = mid + 1;
      } else if (gcdtbl[mid] >= high) {
        right = mid - 1;
      }
    }
    cout << ans << endl;
  }
}