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 mod = 1e9 + 7;
long long C(int n, int r) {
  if (r > n) return 0;
  if (r > n - r) r = n - r;
  long long ans = 1;
  int i;
  for (i = 1; i <= r; i++) {
    ans *= n - r + i;
    ans /= i;
  }
  return ans;
}
vector<long long> sieve(long long n) {
  vector<long long> is_prime(n + 1, true);
  is_prime[0] = is_prime[1] = false;
  for (int i = 2; i <= n; i++) {
    if (is_prime[i] && (long long)i * i <= n) {
      for (int j = i * i; j <= n; j += i) is_prime[j] = false;
    }
  }
  return is_prime;
}
long long fact(int x) {
  long long ans = 1;
  for (long long i = 2; i <= x; ++i) ans *= i;
  return ans;
}
vector<pair<int, int>> factors;
vector<int> divisors;
void get_factors(int n) {
  factors.clear();
  int d = 1;
  for (int i = 2; i * i <= n; i += d, d = 2) {
    if (n % i == 0) {
      factors.push_back({i, 0});
      while (n % i == 0) {
        n /= i;
        factors.back().second++;
      }
    }
  }
  if (n != 1) {
    factors.push_back({n, 1});
  }
}
void get_div(int ind = 0, int res = 1) {
  if (ind == (int)factors.size()) {
    divisors.push_back(res);
    return;
  }
  for (int i = 0; i <= factors[ind].second; ++i) {
    get_div(ind + 1, res);
    res *= factors[ind].first;
  }
}
void test() {
  int a, b;
  cin >> a >> b;
  get_factors(a);
  divisors.clear();
  get_div();
  vector<int> diva = divisors;
  get_factors(b);
  divisors.clear();
  get_div();
  vector<int> divb = divisors;
  sort(diva.begin(), diva.end());
  sort(divb.begin(), divb.end());
  vector<int> cd;
  set_intersection(diva.begin(), diva.end(), divb.begin(), divb.end(),
                   inserter(cd, cd.begin()));
  int t;
  cin >> t;
  for (long long(l) = 0; (l) < t; ++(l)) {
    int low, high;
    cin >> low >> high;
    int ind = upper_bound(cd.begin(), cd.end(), high) - cd.begin();
    ind--;
    if (ind == -1 || cd[ind] < low)
      cout << "-1\n";
    else
      cout << cd[ind] << "\n";
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  for (long long T = 0; T < t; T++) {
    test();
  }
  cerr << "Time : " << 1000 * ((double)clock()) / (double)CLOCKS_PER_SEC
       << "ms\n";
}