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>
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
using namespace std;
const int md = (2 << 30);
const long long oo = 1e14;
const double pi = acos(-1);
int dy[] = {0, 1, 0, -1};
int dx[] = {1, 0, -1, 0};
const int N = 1e4 + 5;
bool pr[N];
void Build() {
  memset(pr, 1, sizeof pr);
  for (int i = 2; i < N; ++i)
    if (pr[i])
      for (int j = i + i; j < N; j += i) pr[j] = 0;
}
bool Check(int x) {
  int cnt = 0, trm = sqrt(x) + 2;
  for (int i = 2; i <= x && cnt < 3; ++i)
    if (!(x % i) && pr[i]) ++cnt;
  return cnt >= 3;
}
int main() {
  int a, b;
  scanf("%d%d", &a, &b);
  deque<int> dq;
  int trm = sqrt(max(a, b)) + 5;
  dq.push_back(1);
  if (!(a % b)) dq.push_back(b);
  if (!(b % a)) dq.push_back(a);
  for (int i = 2; i <= trm; ++i) {
    if (!(a % i) && !(b % i)) dq.push_back(i);
    if (!(b % (a / i)) && !(a % i)) dq.push_back(a / i);
    if (!(a % (b / i)) && !(b % i)) dq.push_back(b / i);
  }
  sort(dq.begin(), dq.end());
  int n, l, r;
  scanf("%d", &n);
  while (n--) {
    scanf("%d%d", &l, &r);
    deque<int>::iterator it = lower_bound(dq.begin(), dq.end(), r);
    if (it == dq.end() || ((*it) > r) && it != dq.begin()) --it;
    l <= (*it) && (*it) <= r ? printf("%d\n", (*it)) : puts("-1");
  }
}