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, q, i, j;
  scanf("%d %d", &a, &b);
  set<int> s1, s2;
  for (i = 1; i * i <= a; i++) {
    if (a % i == 0) {
      s1.insert(i);
      s1.insert(a / i);
    }
  }
  for (i = 1; i * i <= b; i++) {
    if (b % i == 0) {
      int bil1 = i, bil2 = b / i;
      if (s1.count(bil1)) {
        s2.insert(bil1);
      }
      if (s1.count(bil2)) {
        s2.insert(bil2);
      }
    }
  }
  set<int>::iterator it;
  int arr[100000];
  int cnt = 0;
  for (it = s2.begin(); it != s2.end(); it++) {
    arr[cnt] = *it;
    cnt++;
  }
  scanf("%d", &q);
  while (q--) {
    int depan, belakang;
    scanf("%d %d", &depan, &belakang);
    int idx = lower_bound(arr, arr + cnt, belakang) - arr;
    if (idx == cnt) {
      idx--;
    }
    if (arr[idx] > belakang) {
      idx--;
    }
    if (idx < 0) {
      printf("-1\n");
    } else {
      if (arr[idx] < depan) {
        printf("-1\n");
      } else {
        printf("%d\n", arr[idx]);
      }
    }
  }
  return 0;
};