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<int> bla;
set<int> fats;
int gcd(int a, int b) {
  if (!b) return a;
  return gcd(b, a % b);
}
void mult(long long int x) {
  vector<int> add;
  for (set<int>::iterator it = fats.begin(), end = fats.end(); it != end; ++it)
    add.push_back(*it * x);
  for (int i = 0, s = add.size(); i < s; ++i) fats.insert(add[i]);
}
void fat(long long int x) {
  long long int i, c, raiz;
  fats.insert(1);
  c = x;
  while (c % 2 == 0) {
    mult(2);
    fats.insert(2);
    c /= 2;
  }
  i = 3;
  raiz = sqrt(c) + 2;
  while (i < raiz && c > 1) {
    while (c % i == 0) {
      mult(i);
      fats.insert(i);
      c /= i;
    }
    i += 2;
    raiz = sqrt(c) + 2;
  }
  if (c > 1) {
    mult(c);
    fats.insert(c);
  }
  for (set<int>::iterator it = fats.begin(); it != fats.end(); ++it)
    bla.push_back(*it);
}
int main() {
  int a, b, n, c, d;
  scanf("%d%d%d", &a, &b, &n);
  int g = gcd(a, b);
  fat(g);
  bla.push_back(g);
  while (n--) {
    scanf("%d%d", &c, &d);
    vector<int>::iterator low = lower_bound(bla.begin(), bla.end(), c);
    vector<int>::iterator high = lower_bound(bla.begin(), bla.end(), d);
    if (low == bla.end()) {
      printf("-1\n");
      continue;
    } else if (high == bla.end())
      --high;
    else if (*high != d)
      --high;
    if (high >= low)
      printf("%d\n", *high);
    else
      printf("-1\n");
  }
  return 0;
}