#include <bits/stdc++.h>
using namespace std;
int main() {
  int t;
  cin >> t;
  while (t--) {
    long long p, q;
    cin >> p >> q;
    if (p < q)
      cout << p << endl;
    else {
      if (p % q != 0)
        cout << p << endl;
      else {
        long long ans = 0, x;
        bool flag = 0;
        for (int i = 2; i * i <= q; i++) {
          if (q % i == 0) {
            flag = 1;
            x = i;
            int cnt = 0, cnt1 = 0;
            long long q1 = q, p1 = p, p2 = p;
            while (q1 % x == 0) {
              q1 = q1 / x;
              cnt++;
            }
            while (p1 % x == 0) {
              p1 = p1 / x;
              cnt1++;
            }
            long long y = cnt1 - cnt + 1;
            while (y--) p2 = p2 / x;
            ans = max(ans, p2);
            cnt = 0, cnt1 = 0;
            x = q / i;
            q1 = q;
            p1 = p;
            p2 = p;
            while (q1 % x == 0) {
              q1 = q1 / x;
              cnt++;
            }
            while (p1 % x == 0) {
              p1 = p1 / x;
              cnt1++;
            }
            y = cnt1 - cnt + 1;
            while (y--) p2 = p2 / x;
            ans = max(ans, p2);
          }
        }
        if (!flag) {
          while (p % q == 0) {
            p = p / q;
          }
          cout << p << endl;
        } else {
          cout << ans << endl;
        }
      }
    }
  }
}

### Prompt

Your challenge is to write a CPP solution to the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
  int t;
  cin >> t;
  while (t--) {
    long long p, q;
    cin >> p >> q;
    if (p < q)
      cout << p << endl;
    else {
      if (p % q != 0)
        cout << p << endl;
      else {
        long long ans = 0, x;
        bool flag = 0;
        for (int i = 2; i * i <= q; i++) {
          if (q % i == 0) {
            flag = 1;
            x = i;
            int cnt = 0, cnt1 = 0;
            long long q1 = q, p1 = p, p2 = p;
            while (q1 % x == 0) {
              q1 = q1 / x;
              cnt++;
            }
            while (p1 % x == 0) {
              p1 = p1 / x;
              cnt1++;
            }
            long long y = cnt1 - cnt + 1;
            while (y--) p2 = p2 / x;
            ans = max(ans, p2);
            cnt = 0, cnt1 = 0;
            x = q / i;
            q1 = q;
            p1 = p;
            p2 = p;
            while (q1 % x == 0) {
              q1 = q1 / x;
              cnt++;
            }
            while (p1 % x == 0) {
              p1 = p1 / x;
              cnt1++;
            }
            y = cnt1 - cnt + 1;
            while (y--) p2 = p2 / x;
            ans = max(ans, p2);
          }
        }
        if (!flag) {
          while (p % q == 0) {
            p = p / q;
          }
          cout << p << endl;
        } else {
          cout << ans << endl;
        }
      }
    }
  }
}

```