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 gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long x, y, z;
vector<long long> v;
int main() {
  while (scanf("%lld%lld", &x, &y) != EOF) {
    v.clear();
    z = gcd(x, y);
    for (int i = 1; i * i <= z; i++) {
      if (z % i == 0) {
        v.push_back(z / i);
        v.push_back(i);
      }
    }
    sort(v.begin(), v.end());
    int T;
    scanf("%d", &T);
    while (T--) {
      long long e1, e2;
      scanf("%lld%lld", &e1, &e2);
      long long p2 = upper_bound(v.begin(), v.end(), e2) - v.begin() - 1;
      long long res = v[p2];
      if (res < e1)
        printf("-1\n");
      else
        printf("%lld\n", res);
    }
  }
  return 0;
}