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 M = 1e9 + 7;
long long expo(long long a, long long b) {
  if (b == 0) return 1;
  if (b % 2 == 0) {
    long long ret = expo(a, b / 2);
    ret = (ret * ret) % 1000000007;
    return ret;
  } else {
    long long ret = expo(a, b / 2);
    ret = (ret * ret) % 1000000007;
    ret = (ret * a) % 1000000007;
    return ret;
  }
}
bool is_prime(int x) {
  for (int i = 2; i * i <= x; i++)
    if (x % i == 0) return false;
  return true;
}
long long inf = 1e18;
int main() {
  int t;
  long long p, q;
  cin >> t;
  while (t--) {
    cin >> p >> q;
    if (p % q)
      cout << p << endl;
    else {
      set<int> s;
      vector<pair<int, int> > v;
      for (int i = 2; i * i <= q; i++) {
        if (q % i == 0 && is_prime(i)) {
          long long x = q;
          int c = 0;
          while (x % i == 0) {
            c++;
            x /= i;
          }
          v.push_back({i, c});
        }
        if (q % i == 0 && is_prime(q / i)) {
          long long x = q;
          int c = 0;
          while (x % (q / i) == 0) {
            c++;
            x /= (q / i);
          }
          v.push_back({(q / i), c});
        }
      }
      if (is_prime(q)) v.push_back({q, 1});
      set<int>::iterator it = s.begin();
      long long ans = -1;
      for (int i = 0; i < v.size(); i++) {
        long long x = p;
        int c = 0;
        while (x % (v[i].first) == 0) {
          x /= v[i].first;
        }
        x *= pow(v[i].first, v[i].second - 1);
        ans = max(ans, x);
      }
      cout << ans << endl;
    }
  }
}
