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 bsearch(vector<long long int> d, long long int l, long long int h,
                      long long int f, long long int r) {
  long long int mid = (f + r) / 2;
  if (f == r) {
    if (d[mid] <= h && d[mid] >= l)
      return d[mid];
    else
      return -1;
  } else if (d[mid] <= h && d[mid + 1] > h && d[mid] >= l) {
    return d[mid];
  } else if (d[mid] < l && d[mid + 1] > h)
    return -1;
  else if (d[mid + 1] <= h)
    return bsearch(d, l, h, mid + 1, r);
  else
    return bsearch(d, l, h, f, mid - 1);
  return -1;
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long int a, b;
  cin >> a >> b;
  long long int mn = min(a, b);
  long long int mx = max(a, b);
  vector<long long int> d;
  for (long long int i = 2; i <= sqrt(mn); i++) {
    if (mn % i == 0) {
      if (mx % i == 0) d.push_back(i);
      if (mx % (mn / i) == 0) d.push_back(mn / i);
    }
  }
  if (mx % mn == 0) d.push_back(mn);
  d.push_back(1);
  sort(d.begin(), d.end());
  long long int q;
  cin >> q;
  while (q--) {
    long long int l, h;
    cin >> l >> h;
    long long int ans = 0;
    cout << bsearch(d, l, h, 0, d.size() - 1) << endl;
  }
}