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 p, q, x;
map<long long, int> rtq;
void div(map<long long, int>& mp, long long v) {
  for (long long i = 2; i <= sqrt(v); i++) {
    int cnt = 0;
    while (v % i == 0) {
      cnt++;
      v /= i;
    }
    if (cnt != 0) mp[i] = cnt;
  }
  if (v > 1) {
    mp[v] = 1;
  }
}
long long calc(long long v, long long x) {
  long long ret = 1;
  for (long long i = 1; i <= x; i++) ret *= v;
  return ret;
}
long long solve(long long val, map<long long, int>& rt) {
  long long ret = -1;
  for (auto item : rt) {
    long long v = item.first;
    long long val_now = val;
    int cnt = 0;
    while (val_now % v == 0) {
      val_now /= v;
      ++cnt;
    }
    long long div_now = calc(v, cnt - item.second + 1);
    ret = max(ret, val / div_now);
  }
  return ret;
}
int w33ha() {
  rtq.clear();
  scanf("%lld%lld", &p, &q);
  if (p % q != 0) return printf("%lld\n", p);
  div(rtq, q);
  printf("%lld\n", solve(p, rtq));
  return 0;
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) w33ha();
  return 0;
}
