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 d[100005], low[100005], high[100005];
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
int main() {
  int a, b;
  scanf("%d%d", &a, &b);
  int c = gcd(a, b);
  int i = 0, j = 0;
  for (i = 2;; i++) {
    if (i * i > c) break;
    if (c % i == 0) {
      d[j++] = c / i;
      d[j++] = i;
    }
  }
  d[j++] = 1;
  d[j++] = c;
  sort(d, d + j);
  int n;
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    scanf("%d%d", &low[i], &high[i]);
  }
  for (i = 0; i < n; i++) {
    if (c < low[i])
      printf("-1\n");
    else {
      int l = 0, r = j;
      while (l + 1 < r) {
        int mid = (l + r) / 2;
        if (d[mid] > high[i]) {
          r = mid;
        } else {
          l = mid;
        }
      }
      if (d[l] >= low[i])
        printf("%d\n", d[l]);
      else
        printf("-1\n");
    }
  }
  return 0;
}