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> d, d1, d2;
int main() {
  cin >> a >> b >> n;
  for (int i = 1; i * i <= min(a, b); ++i) {
    int x = a / i, y = b / i;
    if (a % i == 0 && b % i == 0) {
      d.push_back(i);
    }
    if (a % i == 0 && a % x == 0 && b % x == 0) {
      d.push_back(x);
    }
    if (b % i == 0 && a % y == 0 && b % y == 0) {
      d.push_back(y);
    }
  }
  sort(d.begin(), d.end());
  while (n--) {
    int l, r, f = 0;
    scanf("%d%d", &l, &r);
    for (int i = d.size() - 1; i >= 0; --i) {
      if (d[i] >= l && d[i] <= r) {
        f = 1;
        printf("%d\n", d[i]);
        break;
      }
    }
    if (f == 0) {
      printf("-1\n");
    }
  }
  return 0;
}