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() {
  long long a, b, tmp, tmp2, l, r, q;
  cin >> a >> b;
  vector<long long> vec, vec2;
  set<long long> st;
  set<long long>::iterator it;
  for (int i = 1; i * i <= a; i++) {
    if (a % i == 0) {
      if (i == a / i)
        vec.push_back(i);
      else {
        vec.push_back(i), vec.push_back(a / i);
      }
    }
  }
  vec.push_back(a);
  for (int i = 1; i * i <= b; i++) {
    if (b % i == 0) {
      if (i == b / i)
        vec2.push_back(i);
      else {
        vec2.push_back(i), vec2.push_back(b / i);
      }
    }
  }
  vec2.push_back(b);
  for (int i = 0; i < vec.size(); i++) {
    for (int j = 0; j < vec2.size(); j++) {
      if (vec[i] == vec2[j]) {
        st.insert(vec[i]);
        break;
      }
    }
  }
  cin >> q;
  while (q--) {
    cin >> l >> r;
    tmp = -1;
    for (it = st.begin(); it != st.end(); it++) {
      if (*it > r)
        break;
      else if (*it >= l && *it <= r)
        tmp = *it;
    }
    cout << tmp << "\n";
  }
  return 0;
}