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 long long INF = 0x7fffffffffffffff;
long long qpow(long long x, long long y) {
  long long ans = 1;
  while (y) {
    if (y & 1) ans = ans * x;
    x = x * x;
    y >>= 1;
  }
  return ans;
}
map<long long, long long> cnt;
int main() {
  int t;
  cin >> t;
  while (t--) {
    long long n, m;
    cin >> n >> m;
    if (n % m != 0)
      cout << n << endl;
    else {
      long long tmp = m;
      for (long long i = 2; i * i <= tmp; i++) {
        if (tmp % i == 0) {
          while (tmp % i == 0) cnt[i]++, tmp /= i;
        }
      }
      if (tmp > 1) cnt[tmp]++;
      long long ans = INF, nn = n;
      for (auto it : cnt) {
        long long x = it.first, y = it.second;
        long long tot = 0;
        while (n % x == 0) tot++, n /= x;
        ans = min(ans, qpow(x, tot - y + 1));
      }
      cout << nn / ans << endl;
    }
    cnt.clear();
  }
  return 0;
}
