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 n, m, i, j, k, s, x, y, sum, T, a[1000005], b[1000005], c[1000005];
int erfen_b(int l, int r, int x) {
  if (l > r) return 0;
  int mid = (l + r) >> 1;
  if (b[mid] == x) return mid;
  if (x < b[mid])
    return erfen_b(l, mid - 1, x);
  else
    return erfen_b(mid + 1, r, x);
}
int erfen_c(int l, int r, int x) {
  if (l > r) return r;
  int mid = (l + r) >> 1;
  if (x >= c[mid])
    return erfen_c(mid + 1, r, x);
  else
    return erfen_c(l, mid - 1, x);
}
int main() {
  cin >> n >> m;
  for (i = 1; i <= (int)(sqrt(n)); i++)
    if (n % i == 0) {
      k++;
      a[k] = i;
      if (n / i != i) {
        k++;
        a[k] = n / i;
      }
    }
  for (i = 1; i <= (int)(sqrt(m)); i++)
    if (m % i == 0) {
      s++;
      b[s] = i;
      if (m / i != i) {
        s++;
        b[s] = m / i;
      }
    }
  sort(a + 1, a + k + 1);
  sort(b + 1, b + s + 1);
  sum = 0;
  for (i = 1; i <= k; i++)
    if (erfen_b(1, s, a[i])) {
      sum++;
      c[sum] = a[i];
    }
  cin >> T;
  while (T--) {
    scanf("%d%d", &x, &y);
    int t = c[erfen_c(1, sum, y)];
    if (t <= y && t >= x) {
      printf("%d\n", t);
    } else
      printf("-1\n");
  }
  return 0;
}