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 MAX_PRIME = 1e6;
int divisors[MAX_PRIME] = {0}, primes[MAX_PRIME] = {0}, nPrimes = 0;
void generatePrimes() {
  for (int i = 2; i < MAX_PRIME; i++) {
    if (divisors[i] == 0) divisors[i] = primes[nPrimes++] = i;
    for (int j = 0; j < nPrimes && i * primes[j] < MAX_PRIME; j++)
      divisors[i * primes[j]] = min(divisors[i], primes[j]);
  }
}
map<long long, int> factorize(long long x) {
  map<long long, int> res;
  for (int i = 0; 1ll * primes[i] * primes[i] <= x; i++) {
    if (x % primes[i] == 0) {
      int c = 0;
      while (x % primes[i] == 0) c++, x /= primes[i];
      res[primes[i]] = c;
    }
  }
  if (x != 1) res[x] = 1;
  return res;
}
int main() {
  generatePrimes();
  int t;
  cin >> t;
  while (t--) {
    long long p;
    int q;
    cin >> p >> q;
    auto f = factorize(q);
    long long res = 1;
    for (auto &[d, c] : f) {
      long long x = p;
      while (x % q == 0) x /= d;
      res = max(res, x);
    }
    cout << res << "\n";
  }
}
