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 bPow(long long a, long long n) {
  if (n == 0) return 1;
  long long p = bPow(a, n / 2);
  p *= p;
  if (n % 2) p *= a;
  return p;
}
signed main() {
  long long t;
  cin >> t;
  for (long long i = 0; i < t; i++) {
    long long p, q;
    cin >> p >> q;
    map<long long, long long> qDivs;
    map<long long, long long> pDivs;
    long long q1 = q;
    long long temp = 2;
    while (q1 != 1) {
      if (temp * temp > q1) {
        qDivs[q1]++;
        break;
      }
      if (q1 % temp) {
        temp++;
        continue;
      }
      q1 /= temp;
      qDivs[temp]++;
    }
    long long p1 = p;
    for (auto& it : qDivs) {
      while (p1 % it.first == 0) {
        p1 /= it.first;
        pDivs[it.first]++;
      }
    }
    if (p1 != 1) {
      pDivs[p1]++;
    }
    long long mx = 0;
    if (p % q) {
      mx = p;
    }
    for (auto& it : qDivs) {
      if (!pDivs[it.first]) continue;
      long long minM = it.second - 1;
      if (it.second == 0) minM = 0;
      long long x = p / bPow(it.first, pDivs[it.first] - minM);
      if (p % x == 0 && x % q) {
        mx = max(mx, x);
      }
    }
    cout << mx << endl;
  }
}
