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;
template <class A, class B>
bool maxi(A &a, const B &b) {
  return (a < b) ? (a = b, true) : false;
};
template <class A, class B>
bool mini(A &a, const B &b) {
  return (a > b) ? (a = b, true) : false;
};
int T;
long long p, q, res = 1;
vector<int> primes;
void Sieve() {
  bitset<31722> sieve;
  for (int i = 2; i <= 31622; ++i) {
    if (sieve[i]) continue;
    primes.push_back(i);
    for (int j = i * i; j <= 31622; j += i) sieve[j] = true;
  }
}
long long power(const long long n, const int k) {
  if (k == 0) return 1;
  long long res = power(n, k >> 1);
  res *= res;
  if (k & 1) res *= n;
  return res;
}
int fact(long long x, const int d) {
  int res = 0;
  while (x % d == 0) {
    ++res;
    x /= d;
  }
  return res;
}
int main(int argc, char *argv[]) {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  Sieve();
  cin >> T;
  int cnt;
  while (T--) {
    cin >> p >> q;
    res = 1;
    for (const int &i : primes) {
      if (q % i == 0) {
        cnt = 0;
        while (q % i == 0) {
          q /= i;
          ++cnt;
        };
        maxi(res, p / power(i, max(0, fact(p, i) - cnt + 1)));
        if (q == 1) break;
      }
    }
    if (q > 1) maxi(res, p / power(q, max(0, fact(p, q))));
    cout << res << '\n';
  }
  return 0;
}
