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;
long long n, a, b, pt_a[100100], pt_b[100100];
vector<long long> tsnt, usnt;
void pt() {
  for (int i = 2; i <= sqrt(a); i++) {
    while (a % i == 0) {
      a /= i;
      pt_a[i]++;
    }
  }
  for (int i = 2; i <= sqrt(b); i++) {
    while (b % i == 0) {
      b /= i;
      pt_b[i]++;
    }
  }
  if (a > 0 && a <= round(1e5)) pt_a[a]++;
  if (b > 0 && b <= round(1e5)) pt_b[b]++;
  for (int i = 2; i <= 100000; i++) {
    long long tmp = min(pt_a[i], pt_b[i]);
    for (int j = 1; j <= tmp; j++) tsnt.push_back(i);
  }
}
void sinh(int i, long long tmp) {
  if (i >= tsnt.size()) {
    usnt.push_back(tmp);
    return;
  }
  sinh(i + 1, tmp);
  sinh(i + 1, tmp * tsnt[i]);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> a >> b;
  cin >> n;
  pt();
  sinh(0, 1);
  sort(usnt.begin(), usnt.end());
  while (n--) {
    int u, v, usmax = 0;
    cin >> u >> v;
    int l = 0;
    int r = usnt.size() - 1;
    while (l <= r) {
      int mid = (l + r) / 2;
      if (usnt[mid] >= u && usnt[mid] <= v) {
        l = mid + 1;
        usmax = max(usmax * 1LL, usnt[mid]);
      } else if (usnt[mid] > v)
        r = mid - 1;
      else if (usnt[mid] < u)
        l = mid + 1;
    }
    if (!usmax)
      cout << "-1\n";
    else
      cout << usmax << "\n";
  }
  return 0;
}