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;
int main() {
  int a, b;
  cin >> a >> b;
  vector<int> gcd;
  if (a > b) {
    int t = a;
    a = b;
    b = t;
  }
  for (int i = 1; i * i <= a; i++) {
    if (a % i == 0 && b % i == 0) gcd.push_back(i);
    if (a % i == 0 && b % (a / i) == 0) gcd.push_back(a / i);
  }
  if (a == b || b % a == 0) gcd.push_back(a);
  sort(gcd.begin(), gcd.end());
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    int low, high;
    bool flag = false;
    scanf("%i%i", &low, &high);
    for (int i = gcd.size() - 1; i >= 0; i--)
      if (gcd[i] >= low && gcd[i] <= high) {
        printf("%i\n", gcd[i]);
        flag = true;
        break;
      }
    if (!flag) printf("-1\n");
  }
  return 0;
}