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> A, B;
long long a, b;
void da(long long n) {
  for (long long i = 1; i * i <= n; i++) {
    if (n % i == 0 && (n / i) != i) {
      A.push_back(i);
      A.push_back(n / i);
    } else if (n % i == 0)
      A.push_back(i);
  }
}
void db(long long n) {
  for (long long i = 1; i * i <= n; i++) {
    if (n % i == 0 && (n / i) != i) {
      B.push_back(i);
      B.push_back(n / i);
    } else if (n % i == 0)
      B.push_back(i);
  }
}
long long ok(long long mid) {
  int i = lower_bound(A.begin(), A.end(), mid) - A.begin();
  int j = lower_bound(B.begin(), B.end(), mid) - B.begin();
  if (i == A.size() || j == B.size() || A[i] > b || B[j] > b) return -1;
  while (i < A.size() && j < B.size() && A[i] <= b && B[j] <= b) {
    if (A[i] == B[j]) return A[i];
    if (A[i] < B[j])
      i = lower_bound(A.begin(), A.end(), B[j]) - A.begin();
    else if (A[i] > B[j])
      j = lower_bound(B.begin(), B.end(), A[i]) - B.begin();
  }
  return -1;
}
long long bs() {
  long long lo = a;
  long long hi = b;
  long long res = -1;
  while (lo <= hi) {
    long long mid = (lo + hi) / 2;
    long long check = ok(mid);
    if (check == -1)
      hi = mid - 1;
    else {
      res = check;
      lo = res + 1;
    }
  }
  return res;
}
int main() {
  cin >> a >> b;
  da(a);
  db(b);
  int n;
  cin >> n;
  sort(A.begin(), A.end());
  sort(B.begin(), B.end());
  for (int i = 0; i < n; i++) {
    cin >> a >> b;
    cout << bs() << endl;
  }
}