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;
const double pi = 3.14159265358979323;
long long int n, k, m, tc, ans = 0, count = 0, mymax = -1e18, mymin = 1e18;
string final_ans = "";
string s = "";
vector<long long int> vec;
unordered_map<int, int> mymap;
unordered_map<char, int> mymap2;
unordered_map<long long int, long long int> prime_factorizer(long long int x) {
  unordered_map<long long int, long long int> res;
  for (long long int i = 2; i <= sqrt(x) + 3; i++) {
    while (x % i == 0) {
      res[i]++;
      x /= i;
    }
  }
  if (x != 1) res[x]++;
  return res;
}
long long int modpow(long long int base, long long int exp) {
  long long int res = 1;
  while (exp >= 1) {
    if (exp % 2) res *= base;
    base *= base;
    exp /= 2;
  }
  return res;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> tc;
  for (int i = 0; i < tc; i += 1) {
    cin >> n >> m;
    long long int orig_n = n;
    if (n % m != 0)
      cout << n << '\n';
    else {
      unordered_map<long long int, long long int> extra = prime_factorizer(m);
      long long int least_divide = 1e18;
      for (auto const& x : extra) {
        long long int base = x.first;
        long long int exp = x.second;
        long long int curr = 0;
        while (n % base == 0) {
          n /= base;
          curr++;
        }
        least_divide = min(least_divide, modpow(base, curr - exp + 1));
      }
      cout << orig_n / least_divide << '\n';
    }
  }
}
