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 int N = 1e9;
int t, isprime[100000];
long long p, q;
vector<long long> v;
vector<pair<long long, long long> > z;
pair<long long, long long> tmp;
void sieve() {
  for (int i = 0; i <= sqrt(N); i++) isprime[i] = true;
  isprime[0] = isprime[1] = false;
  for (int i = 2; i <= sqrt(N); i++) {
    if (isprime[i]) {
      v.push_back(i);
      for (int j = i + i; j <= sqrt(N); j += i) {
        isprime[j] = false;
      }
    }
  }
}
int main() {
  cin >> t;
  sieve();
  while (t--) {
    cin >> p >> q;
    long long go = q;
    z.clear();
    for (int i = 0; i < v.size(); i++) {
      int ok = 0;
      while (q % v[i] == 0) {
        ok++;
        q /= v[i];
      }
      if (ok) {
        z.push_back({v[i], ok});
      }
    }
    if (q != 1) {
      z.push_back({q, 1});
    }
    if (p % go != 0) {
      cout << p << endl;
    } else {
      long long ans = 0;
      for (int i = 0; i < z.size(); i++) {
        tmp = z[i];
        long long lin = p, zz = tmp.first, cnt = tmp.second - 1;
        while (lin % zz == 0) {
          lin /= zz;
        }
        while (cnt--) {
          lin *= zz;
        }
        ans = max(ans, lin);
      }
      cout << ans << endl;
    }
  }
}
