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;
int n;
long long P, Q, p, q, fans;
vector<pair<long long, int> > frac;
long long qpower(long long a, long long x) {
  long long ret = 1;
  if (x <= 0) return 1;
  for (; x; x >>= 1, a = a * a) x& 1 ? ret = ret * a : 0;
  return ret;
}
void solve() {
  scanf("%lld %lld", &p, &Q);
  frac.clear();
  for (long long i = 2; 1ll * i * i <= Q; ++i) {
    if (!(Q % i)) {
      int e = 0;
      while (!(Q % i)) e++, Q /= i;
      frac.push_back(make_pair(i, e));
    }
  }
  if (Q > 1) frac.push_back(make_pair(Q, 1));
  long long tmp = LONG_LONG_MAX;
  for (auto cur : frac) {
    P = p;
    long long num = cur.first;
    int cnt = 0;
    while (!(P % num)) P /= num, cnt++;
    tmp = min(tmp, qpower(num, cnt - cur.second + 1));
  }
  printf("%lld\n", p / tmp);
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) solve();
  return 0;
}
