Oleg's favorite subjects are History and Math, and his favorite branch of mathematics is division.

To improve his division skills, Oleg came up with t pairs of integers p_i and q_i and for each pair decided to find the greatest integer x_i, such that: 

  * p_i is divisible by x_i; 
  * x_i is not divisible by q_i. 

Oleg is really good at division and managed to find all the answers quickly, how about you?

Input

The first line contains an integer t (1 ≤ t ≤ 50) — the number of pairs.

Each of the following t lines contains two integers p_i and q_i (1 ≤ p_i ≤ 10^{18}; 2 ≤ q_i ≤ 10^{9}) — the i-th pair of integers.

Output

Print t integers: the i-th integer is the largest x_i such that p_i is divisible by x_i, but x_i is not divisible by q_i.

One can show that there is always at least one value of x_i satisfying the divisibility conditions for the given constraints.

Example

Input


3
10 4
12 6
179 822


Output


10
4
179

Note

For the first pair, where p_1 = 10 and q_1 = 4, the answer is x_1 = 10, since it is the greatest divisor of 10 and 10 is not divisible by 4.

For the second pair, where p_2 = 12 and q_2 = 6, note that 

  * 12 is not a valid x_2, since 12 is divisible by q_2 = 6; 
  * 6 is not valid x_2 as well: 6 is also divisible by q_2 = 6. 

The next available divisor of p_2 = 12 is 4, which is the answer, since 4 is not divisible by 6.
#include <bits/stdc++.h>
using namespace std;
const int maxn = 3e5 + 5;
const long long mod = 998244353;
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    long long a, b;
    scanf("%lld%lld", &a, &b);
    if (a % b) {
      printf("%lld\n", a);
      continue;
    }
    vector<int> cnt(1e5 + 5);
    vector<int> p;
    int id = 0;
    for (long long i = 2; i * i <= b; ++i) {
      if (b % i == 0) p.push_back(i), ++id;
      while (b % i == 0) b /= i, cnt[id]++;
    }
    if (b != 1) ++id, cnt[id]++, p.push_back(b);
    long long ans = 0;
    for (int i = 0; i < p.size(); ++i) {
      long long pp = p[i];
      long long nump = a;
      long long cntt = 0;
      while (nump % pp == 0) ++cntt, nump /= pp;
      if (cntt < cnt[i + 1])
        continue;
      else {
        while (cnt[i + 1]) {
          nump *= pp;
          --cnt[i + 1];
        }
        nump /= pp;
        ans = max(ans, nump);
      }
    }
    printf("%lld\n", ans);
  }
}
