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 int power(long long int n, long long int m) {
  if (m == 0) return 1;
  long long int ans = power(n, m / 2);
  ans = ans * ans;
  if (m % 2 == 1) ans = ans * n;
  return ans;
}
signed main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  long long int T = 1;
  cin >> T;
  bool p[31624];
  for (long long int i = 2; i <= 31623; ++i) {
    p[i] = 1;
  }
  for (long long int i = 2; i * i <= 31624; ++i) {
    if (p[i]) {
      for (long long int j = i * i; j <= 31623; j += i) {
        p[j] = 0;
      }
    }
  }
  vector<long long int> pr;
  for (long long int i = 2; i <= 31623; ++i) {
    if (p[i]) pr.push_back(i);
  }
  while (T--) {
    long long int a, b;
    cin >> a >> b;
    if (a < b) {
      cout << a << "\n";
      continue;
    }
    if (a % b != 0) {
      cout << a << "\n";
      continue;
    }
    map<long long int, long long int> m1, m2;
    long long int ans = 1e18;
    for (long long int x : pr) {
      while (b % x == 0) {
        m1[x]++;
        b /= x;
      }
    }
    if (b != 1) m1[b]++;
    long long int a1 = a;
    for (auto x : m1) {
      while (a1 % (x.first) == 0) {
        m2[x.first]++;
        a1 /= (x.first);
      }
      ans = min(ans, power(x.first, m2[x.first] - x.second + 1));
    }
    if (b != 1) {
      while (a1 % b == 0) {
        m2[b]++;
        a1 /= b;
      }
      ans = min(ans, power(b, m2[b]));
    }
    cout << a / ans << "\n";
  }
  return 0;
}
