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;
vector<long long> v, v1, v2;
void findDevisor(long long n) {
  long long i = 1;
  while (i * i <= n) {
    if (n % i == 0) {
      v1.push_back(i);
      v1.push_back(n / i);
    }
    i++;
  }
}
void findDevisor1(long long n) {
  long long i = 1;
  while (i * i <= n) {
    if (n % i == 0) {
      v2.push_back(i);
      v2.push_back(n / i);
    }
    i++;
  }
}
long long findans(long long n, long long m) {
  for (int i = v.size() - 1; i >= 0; i--) {
    if (v[i] >= n && v[i] <= m) return v[i];
  }
  return -1;
}
int main() {
  long long a, b;
  cin >> a >> b;
  findDevisor(a);
  findDevisor1(b);
  sort(v1.begin(), v1.end());
  sort(v2.begin(), v2.end());
  int i = 0, j = 0;
  while (1) {
    if (i == v1.size() || v2.size() == j) break;
    if (v1[i] == v2[j]) {
      v.push_back(v1[i]);
      i++;
      j++;
    } else if (v1[i] < v2[j])
      i++;
    else
      j++;
  }
  int ln = v.size();
  int n;
  for (cin >> n; n; n--) {
    long long x, y;
    cin >> x >> y;
    cout << findans(x, y) << endl;
  }
  return 0;
}