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;
const int maxn = 40000 + 10;
int a, b, n, gcd, low, hig;
int arr[maxn * 2], cnt;
int Gcd(int a, int b) {
  if (b == 0) return a;
  return Gcd(b, a % b);
}
void GetYue(int tar) {
  for (int i = 1; i * i <= tar; i++) {
    if (tar % i == 0) {
      arr[cnt++] = i;
      if (tar / i != i) arr[cnt++] = tar / i;
    }
  }
  sort(arr, arr + cnt);
}
int Query(int lo, int hi) {
  int p = upper_bound(arr, arr + cnt, hi) - arr - 1;
  return arr[p] >= lo ? arr[p] : -1;
}
int main() {
  while (scanf("%d%d", &a, &b) != EOF) {
    scanf("%d", &n);
    gcd = Gcd(a, b);
    GetYue(gcd);
    for (int i = 0; i < n; i++) {
      scanf("%d%d", &low, &hig);
      printf("%d\n", Query(low, hig));
    }
  }
  return 0;
}