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 gcd(int a, int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
inline int bsr(int l, int r, int d) {
  int res = -1, mid;
  while (l <= r) {
    mid = (l + r) / 2;
    if (d % mid == 0) {
      res = mid;
      l = mid + 1;
    } else if (d % mid <= d / mid)
      l = mid + 1;
    else
      r = mid - 1;
  }
  return res;
}
int main() {
  int a, b, d, l, r;
  scanf("%d%d", &a, &b);
  d = gcd(a, b);
  vector<int> div;
  for (int i = 1; i * i <= d; i++) {
    if (d % i == 0) {
      div.push_back(i);
      if (i != d / i) div.push_back(d / i);
    }
  }
  sort(div.rbegin(), div.rend());
  int n, res;
  scanf("%d", &n);
  while (n--) {
    scanf("%d%d", &l, &r);
    res = -1;
    for (int i = 0; i < div.size(); i++) {
      if (div[i] >= l && div[i] <= r) {
        res = div[i];
        break;
      }
    }
    printf("%d\n", res);
  }
  return 0;
}