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 gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
void solve() {
  long long p, q;
  cin >> p >> q;
  long long g = p;
  long long t = q;
  map<long long, long long> f, l;
  for (long long i = 2; i * i <= t; i++) {
    if (t % i == 0) {
      while (g % i == 0) {
        g /= i;
        f[i]++;
      }
      while (t % i == 0) {
        t /= i;
        l[i]++;
      }
    }
  }
  if (t != 1) {
    while (g % t == 0) {
      g /= t;
      f[t]++;
    }
    l[t]++;
    t = 1;
  }
  long long ans = g;
  for (auto x : l) {
    long long u = p;
    for (int i = 0; i < max(0ll, f[x.first] - x.second + 1); i++) u /= x.first;
    ans = max(ans, u);
  }
  cout << ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int q = 1;
  cin >> q;
  for (; q > 0; q--) {
    solve();
    cout << endl;
  }
}
