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 a, b, n, from, to;
vector<pair<int, int> > getFactors(int n) {
  vector<pair<int, int> > ret;
  for (int k = 2; k * k <= n; ++k) {
    if (n % k == 0) {
      ret.push_back(make_pair(k, 0));
      while (n % k == 0) {
        n /= k;
        ++ret.back().second;
      }
    }
  }
  if (n != 1) {
    ret.push_back(make_pair(n, 1));
  }
  return (ret);
}
vector<int> getDivisors(vector<pair<int, int> > &factors, int i, int n) {
  if (i == factors.size()) {
    vector<int> ret;
    ret.push_back(n);
    return (ret);
  }
  vector<int> ret;
  for (int j = 0; j <= factors[i].second; ++j) {
    vector<int> tmp = getDivisors(factors, i + 1, n);
    for (int k = 0; k < tmp.size(); ++k) {
      ret.push_back(tmp[k]);
    }
    n *= factors[i].first;
  }
  return (ret);
}
int main() {
  scanf("%d %d %d", &a, &b, &n);
  vector<pair<int, int> > factA = getFactors(a);
  vector<pair<int, int> > factB = getFactors(b);
  vector<int> divA = getDivisors(factA, 0, 1);
  vector<int> divB = getDivisors(factB, 0, 1);
  vector<int> divC;
  sort(divA.begin(), divA.end());
  sort(divB.begin(), divB.end());
  set_intersection(divA.begin(), divA.end(), divB.begin(), divB.end(),
                   back_inserter(divC));
  for (int i = 1; i <= n; ++i) {
    scanf("%d %d", &from, &to);
    bool fl = false;
    for (int j = divC.size() - 1; j >= 0 && !fl; --j) {
      if (divC[j] >= from && divC[j] <= to) {
        printf("%d\n", divC[j]);
        fl = true;
      }
    }
    if (!fl) {
      printf("-1\n");
    }
  }
  return 0;
}