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>
inline long long min(const long long &a, const long long &b) {
  return a < b ? a : b;
}
inline long long max(const long long &a, const long long &b) {
  return a > b ? a : b;
}
long long p[100010], l, cnt[100010];
void get() {
  l = 0;
  long long a, b, i;
  scanf("%lld%lld", &a, &b);
  if (a < b || a % b) {
    printf("%lld\n", a);
    return;
  }
  for (i = 2; i * i <= b; i++) {
    if (b % i == 0) {
      p[++l] = i;
      cnt[l] = 0;
      while (b % i == 0) b /= i, cnt[l]++;
    }
  }
  if (b != 1) p[++l] = b, cnt[l] = 1;
  long long count, cp, ans = 0;
  for (i = 1; i <= l; i++) {
    count = 0, cp = a;
    while (cp % p[i] == 0) cp /= p[i], count++;
    count = min(count, cnt[i] - 1);
    if (count > 0)
      while (count--) cp *= p[i];
    ans = max(ans, cp);
  }
  printf("%lld\n", ans);
}
signed main() {
  long long T;
  scanf("%lld", &T);
  while (T--) get();
  return 0;
}
