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 fastPower(long long b, long long p) {
  if (!p) return 1;
  long long answer = fastPower(b, p / 2);
  answer = ((answer % (long long)(1e9 + 7)) * (answer % (long long)(1e9 + 7))) %
           (long long)(1e9 + 7);
  if (p % 2)
    answer = ((answer % (long long)(1e9 + 7)) * (b % (long long)(1e9 + 7))) %
             (long long)(1e9 + 7);
  return answer;
}
vector<long long> divisors;
long long GCD(long long a, long long b) {
  if (!b) return a;
  return GCD(b, a % b);
}
void Div(long long num) {
  long long max = sqrt(num);
  for (int i = 1; i <= max; i++) {
    if (num % i == 0) {
      long long res = num / i;
      divisors.push_back(i);
      if (res != i) divisors.push_back(res);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long a, b;
  cin >> a >> b;
  long long gcd = GCD(a, b);
  Div(gcd);
  sort(divisors.begin(), divisors.end());
  int t;
  cin >> t;
  while (t--) {
    long long l, r;
    cin >> l >> r;
    long long maxGCD = -1;
    for (auto i : divisors) {
      if (i >= l && i <= r) maxGCD = i;
    }
    cout << maxGCD << endl;
  }
}