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 m = 1000000007;
long long fact(long long n);
int modInverse(int a, int m) {
  int m0 = m;
  int y = 0, x = 1;
  if (m == 1) return 0;
  while (a > 1) {
    int q = a / m;
    int t = m;
    m = a % m, a = t;
    t = y;
    y = x - q * y;
    x = t;
  }
  if (x < 0) x += m0;
  return x;
}
long long fact(long long n) {
  long long res = 1;
  for (long long i = 2; i <= n; i++) res = (res % m * i % m) % m;
  return res;
}
int main() {
  long long t;
  cin >> t;
  while (t--) {
    long long p, q;
    cin >> p >> q;
    long long s = 1;
    if (p < q) {
      cout << p << endl;
      continue;
    }
    if (p % q) {
      cout << p << endl;
      continue;
    }
    vector<long long> v;
    v.push_back(q);
    for (long long i = 2; i * i <= q; i++) {
      if (q % i == 0) {
        v.push_back(i);
        if (i != q / i) v.push_back(q / i);
      }
    }
    for (long long i = 0; i < v.size(); i++) {
      long long x = p;
      while (x % v[i] == 0) {
        x = x / v[i];
        if (x % q != 0) {
          s = max(s, x);
          break;
        }
      }
    }
    cout << s << endl;
  }
  return 0;
}
