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;
const int maxn = 2e5 + 5;
map<int, int> mp;
void solve(int x) {
  for (int i = 2; i <= x / i; i++) {
    int cnt = 0;
    if (x % i == 0) {
      while (x % i == 0) x /= i, cnt++;
      mp[i] = cnt;
    }
  }
  if (x != 1) {
    mp[x] = 1;
  }
}
long long pow(int x, int y) {
  long long ans = 1;
  while (y) {
    if (y & 1) ans = ans * x;
    x = x * x;
    y /= 2;
  }
  return ans;
}
long long ask(long long n, long long m) {
  while (n % m == 0) {
    n /= m;
  }
  return n;
}
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    mp.clear();
    long long n;
    int m;
    scanf("%lld %d", &n, &m);
    if (n % m != 0) {
      printf("%lld\n", n);
    } else {
      solve(m);
      long long ans = 0;
      map<int, int>::iterator it;
      for (it = mp.begin(); it != mp.end(); it++) {
        int x = it->first, y = it->second;
        long long tmp = n, tmp1 = 1;
        while (tmp % x == 0) tmp /= x;
        tmp1 = tmp;
        for (int i = 1; i < y; i++) {
          tmp1 *= x;
        }
        ans = max(ans, tmp1);
      }
      printf("%lld\n", ans);
    }
  }
}
