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 warning(disable : 4996)
using namespace std;
int main() {
  int a, b, n;
  scanf("%d%d%d", &a, &b, &n);
  vector<int> res;
  int MIN = min(a, b);
  int MAX = max(a, b);
  set<int> st;
  if (MAX % MIN == 0) {
    st.insert(MIN);
    res.push_back(MIN);
  }
  for (int i = 1; i * i <= MAX; i++) {
    if (MIN % i == 0 && MAX % i == 0) {
      if (st.find(i) == st.end()) {
        res.push_back(i);
        st.insert(i);
      }
    }
    int t = MAX / i;
    if (t != 0 && MIN % t == 0 && MAX % t == 0)
      if (st.find(t) == st.end()) {
        res.push_back(t);
        st.insert(t);
      }
    t = MIN / i;
    if (t != 0 && MIN % t == 0 && MAX % t == 0)
      if (st.find(t) == st.end()) {
        res.push_back(t);
        st.insert(t);
      }
  }
  sort(res.begin(), res.end());
  for (int h = 0; h < n; h++) {
    int low, high;
    scanf("%d%d", &low, &high);
    if (low == high) {
      if (MIN % low == 0 && MAX % high == 0) {
        printf("%d\n", low);
      } else
        printf("-1\n");
      continue;
    }
    int m = lower_bound(res.begin(), res.end(), high) - res.begin();
    if (m + 1 < (int)res.size() && res[m + 1] >= low && res[m + 1] <= high) {
      printf("%d\n", res[m + 1]);
      continue;
    }
    if (m != (int)res.size() && res[m] >= low && res[m] <= high) {
      printf("%d\n", res[m]);
      continue;
    }
    if (m - 1 >= 0 && res[m - 1] >= low && res[m - 1] <= high) {
      printf("%d\n", res[m - 1]);
      continue;
    }
    printf("-1\n");
  }
  return 0;
}