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;
using ll = long long;
int tc;
ll a, b, INF = 1e9;
int main() {
  scanf("%d", &tc);
  while (tc--) {
    scanf("%lld %lld", &a, &b);
    if (a % b) {
      printf("%lld\n", a);
      continue;
    }
    ll A = a, B = b;
    ll mx = 0;
    for (ll i = 2; i * i <= INF; i++) {
      if (i == 1) break;
      if (b % i == 0) {
        int c1 = 0, c2 = 0;
        while (b % i == 0) b /= i, c1++;
        while (a % i == 0) a /= i, c2++;
        int c = c2 - (c1 - 1);
        ll g = 1;
        for (int j = 1; j <= c; j++) g *= i;
        mx = max(mx, A / g);
      }
    }
    if (b > 1) {
      int c1 = 1, c2 = 0;
      while (a % b == 0) a /= b, c2++;
      int c = c2 - (c1 - 1);
      ll g = 1;
      for (int j = 1; j <= c; j++) g *= b;
      mx = max(mx, A / g);
    }
    printf("%lld\n", mx);
  }
}
