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;
vector<long long> sameDivisors(vector<long long> divA, long long sizeA,
                               vector<long long> divB, long long sizeB) {
  vector<long long> intersect;
  if (sizeA < sizeB) {
    for (int i = 0; i < sizeA; i++) {
      if (binary_search(divB.begin(), divB.end(), divA[i])) {
        intersect.push_back(divA[i]);
      }
    }
  } else {
    for (int i = 0; i < sizeB; i++) {
      if (binary_search(divA.begin(), divA.end(), divB[i])) {
        intersect.push_back(divB[i]);
      }
    }
  }
  return intersect;
}
vector<long long> divisors(long long number) {
  vector<long long> d;
  for (long long i = 1; i <= sqrt(number); i++) {
    if (number % i == 0) {
      d.push_back(i);
      if (number / i != i) {
        d.push_back(number / i);
      }
    }
  }
  return d;
}
int main() {
  long long a, b;
  int q;
  cin >> a >> b >> q;
  vector<long long> divA, divB;
  divA = divisors(a);
  sort(divA.begin(), divA.end());
  divB = divisors(b);
  sort(divB.begin(), divB.end());
  vector<long long> div;
  div = sameDivisors(divA, divA.size(), divB, divB.size());
  for (int i = 0; i < q; i++) {
    long long low, high;
    cin >> low >> high;
    bool flag;
    flag = false;
    for (int i = div.size() - 1; i >= 0; i--) {
      if (div[i] >= low && div[i] <= high) {
        cout << div[i] << endl;
        flag = true;
        break;
      }
    }
    if (!flag) {
      cout << "-1" << endl;
    }
  }
  return 0;
}