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;
template <typename Arg1>
void pn(Arg1&& arg1) {
  cout << arg1 << "\n";
}
template <typename Arg1>
void ps(Arg1&& arg1) {
  cout << arg1 << " ";
}
long long powmod(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long modInverse(long long n, long long p) { return powmod(n, p - 2, p); }
void solve() {
  long long n, y, z, p, x, k, l, sum = 0, flag = 0, t, ans = 0;
  string s, s1 = "", s2;
  long long q;
  cin >> p >> q;
  if ((p % q) != 0) {
    pn(p);
    return;
  }
  map<long long, long long> a;
  k = q;
  for (int i = 2; i * i <= q; ++i) {
    t = 0;
    while (k % i == 0) {
      k /= i;
      t++;
    }
    if (t > 0) a[i] = t;
  }
  if (k > 1) a[k] = 1;
  sum = LLONG_MAX;
  for (auto i : a) {
    x = pow(i.first, i.second), t = 0, k = p;
    while ((k % x == 0)) {
      k /= x;
      t++;
    }
    t--;
    l = 0;
    while ((k % (i.first)) == 0) {
      k /= i.first;
      l++;
    }
    l++;
    t = pow(x, t) * pow(i.first, l);
    sum = min(sum, t);
  }
  pn(max(1ll, p / sum));
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int test = 1;
  cin >> test;
  while (test--) {
    solve();
  }
  return 0;
}
