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 gcd(long long a, long long b) {
  while (a > 0 && b > 0) {
    if (a > b)
      a = a % b;
    else
      b = b % a;
  }
  return a + b;
}
void solve() {
  long long a, b;
  cin >> a >> b;
  if (a % b != 0) {
    cout << a << endl;
    return;
  }
  map<long long, long long> d1, d2;
  map<long long, long long>::iterator ii;
  long long rest = a / b;
  long long i;
  vector<long long> prime;
  for (i = 2; i * i <= b; i++) {
    while (b % i == 0) {
      prime.push_back(i);
      b /= i;
      d1[i]++;
    }
  }
  if (b > 1) {
    prime.push_back(b);
    d1[b]++;
  }
  long long t = prime.size();
  for (i = 0; i < t; i++) {
    while (rest % prime[i] == 0) {
      prime.push_back(prime[i]);
      rest /= prime[i];
    }
  }
  for (i = 0; i < prime.size(); i++) {
    d2[prime[i]]++;
  }
  long long total = 1;
  for (i = 0; i < prime.size(); i++) total *= prime[i];
  long long ans = 1e18;
  for (ii = d1.begin(); ii != d1.end(); ii++) {
    long long cur = ii->first, num = ii->second;
    long long diff = d2[cur] - num + 1;
    long long sr = 1;
    for (i = 0; i < diff; i++) sr *= cur;
    ans = min(ans, sr);
  }
  cout << rest * total / ans << endl;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  while (n--) solve();
  return 0;
}
