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>
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("O3")
using namespace std;
const long long N = 2e5;
const long long M = 998244353;
long long bin_pow(long long a, long long n) {
  if (n == 0) return 1;
  if (n % 2 == 1)
    return a * bin_pow(a, n - 1);
  else {
    long long x = bin_pow(a, n / 2);
    return x * x;
  }
}
long long solve() {
  long long p, q;
  cin >> p >> q;
  if (p % q != 0) return p;
  long long k = p;
  for (long long b = 2; b * b <= q; b++) {
    if (q % b != 0) continue;
    long long gg = 0;
    while (q % b == 0) {
      gg++;
      q /= b;
    }
    long long f = 0;
    long long p1 = p;
    while (p1 % b == 0) {
      f++;
      p1 /= b;
    }
    long long st = f - gg + 1;
    k = min(k, bin_pow(b, st));
  }
  if (q > 1) {
    long long b = q;
    long long gg = 0;
    while (q % b == 0) {
      gg++;
      q /= b;
    }
    long long f = 0;
    long long p1 = p;
    while (p1 % b == 0) {
      f++;
      p1 /= b;
    }
    long long st = f - gg + 1;
    k = min(k, bin_pow(b, st));
  }
  return p / k;
}
int32_t main() {
  long long tt;
  cin >> tt;
  while (tt--) {
    cout << (solve()) << "\n";
  }
}
