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;
long long int MOD = 1e9 + 7;
long long int oo = 1e18 + 5;
static int N = 1e5 + 5;
vector<bool> is_prime(N, true);
void sieve() {
  is_prime[0] = is_prime[1] = false;
  for (int i = 2; i * i < N; i++) {
    if (is_prime[i]) {
      for (int j = i * i; j < N; j += i) {
        is_prime[j] = false;
      }
    }
  }
}
int main() {
  sieve();
  int t;
  cin >> t;
  while (t--) {
    long long int p, q;
    cin >> p >> q;
    long long int pb = p;
    if (p % q != 0) {
      cout << p << endl;
      continue;
    }
    map<long long int, int> f_q;
    for (int i = 2; i < N; i++) {
      if (!is_prime[i]) continue;
      if (i * i > q) break;
      while (q % i == 0) {
        f_q[i]++;
        q /= i;
      }
    }
    if (q > 1) {
      f_q[q]++;
    };
    long long int ans = 0;
    for (auto i : f_q) {
      long long int fac = i.first, tm = i.second;
      long long int tmp = p;
      while (tmp % fac == 0) {
        tmp /= fac;
      }
      for (int i = 0; i < tm - 1; i++) {
        tmp *= fac;
      }
      ans = max(ans, tmp);
    }
    cout << ans << endl;
  }
}
