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;
const long long Mod = 1000000007;
void div(int n, vector<int> &divisors) {
  for (int i = 1; i * i <= n; i++) {
    if (n % i == 0) {
      divisors.push_back(i);
      int v = n / i;
      if (v * i == n && n % v == 0) divisors.push_back(v);
    }
  }
}
int main() {
  int n, m;
  scanf("%d %d", &n, &m);
  vector<int> divMin, divMax;
  div(min(n, m), divMin);
  div(max(n, m), divMax);
  sort(divMin.begin(), divMin.end());
  sort(divMax.begin(), divMax.end());
  int q;
  scanf("%d", &q);
  while (q--) {
    int ans = 0;
    int low, high;
    scanf("%d %d", &low, &high);
    for (int i = divMin.size() - 1; i >= 0; i--) {
      int d = divMin[i];
      if (d >= low && d <= high &&
          binary_search(divMax.begin(), divMax.end(), d)) {
        ans = max(ans, d);
        break;
      }
    }
    if (!ans)
      puts("-1");
    else
      printf("%d\n", ans);
  }
  return 0;
}