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 div1[500000], n;
int bin(int low, int high) {
  int i, j, k, l;
  int beg = 0, end = n;
  int mid;
  while (beg <= end) {
    if (div1[beg] > high || div1[end] < low) return -1;
    mid = beg + end;
    mid /= 2;
    if (div1[mid] >= low && div1[mid] <= high) {
      beg = mid;
      if (div1[mid] <= div1[mid + 1]) {
        for (i = mid + 1; i <= end; i++) {
          if (div1[beg] == div1[i]) {
            beg = i;
          } else if (div1[i] <= high) {
            beg = i;
          } else
            break;
        }
        return div1[beg];
      }
    } else if (div1[mid] < low) {
      beg = mid + 1;
    } else if (div1[mid] > high) {
      end = mid - 1;
    }
    if (beg == end && div1[end] >= low && div1[end] <= high)
      return div1[end];
    else if (beg == end)
      return -1;
  }
  return -1;
}
int main() {
  int i, j, k, l, m, n1, n2;
  int flag;
  double t1, t2;
  scanf("%d %d", &n1, &n2);
  if (n1 < n2) {
    k = n1;
    n1 = n2;
    n2 = k;
  }
  t2 = n1;
  t1 = sqrt(t2);
  j = t1;
  k = 0;
  for (i = 1; i <= j; i++) {
    if (n1 % i == 0 & n2 % i == 0) {
      div1[k++] = i;
      if ((n2 % (n1 / i)) == 0) div1[k++] = n1 / i;
      if ((n1 % (n2 / i)) == 0) div1[k++] = n2 / i;
    } else if (n2 % i == 0) {
      if ((n1 % (n2 / i)) == 0) div1[k++] = n2 / i;
    } else if (n1 % i == 0) {
      if ((n2 % (n1 / i)) == 0) div1[k++] = n1 / i;
    }
  }
  n = k - 1;
  sort(div1, div1 + k);
  int q;
  scanf("%d", &q);
  for (i = 1; i <= q; i++) {
    scanf("%d %d", &k, &l);
    printf("%d\n", bin(k, l));
  }
  return 0;
}