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 a, b, n;
vector<int> dvs;
int main() {
  scanf("%d%d%d", &a, &b, &n);
  if (a > b) swap(a, b);
  for (int i = 1; i <= sqrt(a); ++i) {
    if (i * (a / i) == a) {
      if (b % i == 0) dvs.push_back(i);
      if (i != (a / i) && b % (a / i) == 0) dvs.push_back(a / i);
    }
  }
  dvs.push_back(1);
  sort(dvs.begin(), dvs.end());
  while (n--) {
    int l, r;
    scanf("%d%d", &l, &r);
    auto k = upper_bound(dvs.begin(), dvs.end(), r) - dvs.begin();
    --k;
    if (dvs[k] >= l && dvs[k] <= r)
      printf("%d\n", dvs[k]);
    else
      printf("-1\n");
  }
}