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 x, y;
long long T;
long long cnt;
long long vis[100005];
long long pr[100005];
void pre(int lim) {
  for (int i = 2; i <= lim; i++) {
    if (!vis[i]) {
      vis[i] = 1;
      pr[++cnt] = i;
      for (int j = i; j <= lim / i; j++) vis[i * j] = 1;
    }
  }
}
int main() {
  pre(100000);
  scanf("%lld", &T);
  while (T--) {
    scanf("%lld%lld", &x, &y);
    if (x % y != 0) {
      printf("%lld\n", x);
      continue;
    }
    long long ans = 1, tmp = y;
    for (int i = 1; i <= cnt; i++) {
      long long cs = 0, res = 1;
      while (tmp % pr[i] == 0) tmp /= pr[i], cs++, res *= pr[i];
      if (!cs) continue;
      res /= pr[i];
      if (x % res == 0) {
        long long tmp1 = x;
        while (tmp1 % pr[i] == 0) tmp1 /= pr[i];
        ans = max(ans, res * tmp1);
      }
    }
    if (tmp > 1) {
      long long res = 1;
      if (x % res == 0) {
        long long tmp1 = x;
        while (tmp1 % tmp == 0) tmp1 /= tmp;
        ans = max(ans, res * tmp1);
      }
    }
    printf("%lld\n", ans);
  }
  return 0;
}
