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 a, b, ma[100001], mb[100001], ca = 0, cb = 0, g[100001], cnt = 0;
int BS(int L, int R, int x) {
  if (R - L <= 1) {
    if (x >= g[R]) return g[R];
    if (x >= g[L]) return g[L];
    if (L > 0) return g[L - 1];
    return -1;
  }
  int tg = (L + R) / 2;
  if (x == g[tg]) return g[tg];
  if (x < g[tg]) {
    return BS(L, tg - 1, x);
  }
  return BS(tg + 1, R, x);
}
int main() {
  int i, j, n, t1, t2;
  cin >> a >> b;
  ma[ca] = 1;
  ca++;
  for (i = 2; i * i <= a; i++) {
    if (a % i == 0) {
      ma[ca] = i;
      ca++;
    }
  }
  int ta = ca - 1, tb;
  for (i = ta; i >= 0; i--) {
    if (a / ma[i] != ma[i]) {
      ma[ca] = a / ma[i];
      ca++;
    }
  }
  if (a > 1) {
    ma[ca] = a;
    ca++;
  }
  mb[cb] = 1;
  cb++;
  for (i = 2; i * i <= b; i++) {
    if (b % i == 0) {
      mb[cb] = i;
      cb++;
    }
  }
  tb = cb - 1;
  for (i = tb; i >= 0; i--) {
    if (b / mb[i] != mb[i]) {
      mb[cb] = b / mb[i];
      cb++;
    }
  }
  if (b > 1) {
    mb[cb] = b;
    cb++;
  }
  i = 0;
  j = 0;
  while (i < ca && j < cb) {
    if (ma[i] < mb[j]) {
      i++;
    } else if (ma[i] > mb[j]) {
      j++;
    } else {
      g[cnt] = ma[i];
      cnt++;
      i++;
      j++;
    }
  }
  cnt--;
  cin >> n;
  int p, q;
  for (i = 0; i < n; i++) {
    cin >> t1 >> t2;
    p = BS(0, cnt, t2);
    if (p < t1) {
      p = -1;
    }
    if (i > 0)
      cout << endl << p;
    else
      cout << p;
  }
}