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;
int ROW[] = {+0, +0, -1, +1};
int COL[] = {+1, -1, +0, +0};
int X[] = {+0, +0, +1, -1, -1, +1, -1, +1};
int Y[] = {-1, +1, +0, +0, +1, +1, -1, -1};
int KX[] = {-2, -2, -1, -1, 1, 1, 2, 2};
int KY[] = {-1, 1, -2, 2, -2, 2, -1, 1};
long long calc(long long num, long long val, long long cnt1, long long cnt2) {
  while (cnt1 >= cnt2) {
    num /= val;
    cnt1--;
  }
  return num;
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    long long p, q, n, m;
    scanf("%lld %lld", &p, &q);
    n = p;
    m = q;
    if (p % q) {
      printf("%lld\n", p);
      continue;
    }
    map<long long, long long> mp, mpp;
    for (long long i = 2; i * i <= q; i++) {
      if (q % i) continue;
      while (q % i == 0) {
        mp[i]++;
        q /= i;
      }
    }
    if (q > 1) mp[q]++;
    long long ans = 1LL;
    for (auto it : mp) {
      while (p % it.first == 0) {
        p /= it.first;
        mpp[it.first]++;
      }
      ans = max(ans, calc(n, it.first, mpp[it.first], it.second));
    }
    printf("%lld\n", ans);
  }
  return 0;
}
