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;
vector<pair<int, int> > primeFactorise(long long n) {
  vector<pair<int, int> > primeFactors;
  for (long long i = 2; i * i <= n; i++) {
    long long power = 0;
    while (n % i == 0) {
      power++;
      n /= i;
    }
    if (power > 0) primeFactors.push_back({i, power});
  }
  if (n > 1) primeFactors.push_back({n, 1});
  return primeFactors;
}
long long exp(long long b, long long e) {
  long long ans = 1;
  while (e) {
    if (e & 1) ans *= b;
    b = b * b;
    e >>= 1;
  }
  return ans;
}
void solve() {
  long long p, q;
  cin >> p >> q;
  if (p % q != 0) {
    printf("%lld\n", p);
    return;
  }
  vector<pair<int, int> > pf = primeFactorise(q);
  long long best = 1;
  for (int i = 0; i < (int)pf.size(); i++) {
    int a = 0;
    long long n = p;
    while (n % pf[i].first == 0) {
      n /= pf[i].first;
      a++;
    }
    int b = pf[i].second;
    long long factor = exp(pf[i].first, a - (b - 1));
    best = max(best, p / factor);
  }
  printf("%lld\n", best);
}
int main() {
  int t = 0;
  cin >> t;
  for (int i = 0; i < t; i++) solve();
}
