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 v[1000000], i = 0;
int f_mcd(int a, int b) {
  if (b == 0 || a == b)
    return a;
  else
    return f_mcd(b, a % b);
}
void divid(int x) {
  for (int d = 1; d * d <= x; d++) {
    if (x % d == 0) {
      v[i] = d;
      if (x / d != d) {
        v[i + 1] = x / d;
      }
      if (x / d == d) i -= 1;
      i += 2;
    }
  }
  sort(v, v + i);
}
int main() {
  int n, mcd, a, b;
  cin >> a >> b;
  mcd = f_mcd(a, b);
  divid(mcd);
  cin >> n;
  while (n--) {
    int bas, alt;
    cin >> bas >> alt;
    int mi, lo = 0, hi = i - 1;
    while (lo < hi) {
      mi = lo + (hi - lo + 1) / 2;
      if (v[mi] < bas) lo = mi + 1;
      if (v[mi] > alt) hi = mi - 1;
      if (v[mi] <= alt and v[mi] >= bas) {
        lo = mi;
      }
    }
    if (v[lo] <= alt and v[lo] >= bas) {
      cout << v[lo] << '\n';
    } else
      cout << "-1" << '\n';
  }
  return 0;
}