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;
long long int sqr(long long int x) { return x * x; }
inline long long int add(long long int a, long long int b) {
  long long int c = a + b;
  if (c >= 1000000007) c -= 1000000007;
  return c;
}
inline long long int mul(long long int a, long long int b) {
  return (a * 1ll * b) % 1000000007;
}
inline long long int sub(long long int a, long long int b) {
  long long int c = a - b;
  if (c < 0) c += 1000000007;
  return c;
}
const int nax = 200 * 1000 + 10;
void get(int n, vector<int>& div) {
  for (int i = 1; i * i <= n; ++i) {
    if (n % i == 0) {
      div.push_back(i);
      if (i * i != n) {
        div.push_back(n / i);
      }
    }
  }
}
int main() {
  int a, b;
  scanf("%d %d", &a, &b);
  vector<int> fa;
  get(a, fa);
  vector<int> fb;
  get(b, fb);
  sort(fa.begin(), fa.end());
  sort(fb.begin(), fb.end());
  vector<int> ans;
  set_intersection(fa.begin(), fa.end(), fb.begin(), fb.end(),
                   inserter(ans, ans.begin()));
  int q;
  scanf("%d", &q);
  while (q--) {
    int l, h;
    scanf("%d %d", &l, &h);
    int id = upper_bound(ans.begin(), ans.end(), h) - ans.begin() - 1;
    if (id == -1) {
      printf("%d\n", -1);
    } else if (ans[id] >= l && ans[id] <= h) {
      printf("%d\n", ans[id]);
    } else {
      printf("%d\n", -1);
    }
  }
  return 0;
}