#include <bits/stdc++.h>
using namespace std;
bool checkPrime(long long n) {
  long long cnt = 0;
  for (long long i = 2; i <= sqrt(n); i++) {
    if ((n % i) == 0) {
      return false;
    }
  }
  return true;
}
int main() {
  long long t;
  cin >> t;
  while (t--) {
    long long a, b;
    cin >> a >> b;
    if ((a % b) != 0) {
      cout << a << endl;
    } else {
      long long a1, a2;
      bool bb;
      unordered_map<long long, long long> ma;
      unordered_map<long long, long long> mb;
      unordered_map<long long, long long>::iterator itr;
      for (long long j = 1; j <= sqrt(b); j++) {
        a1 = j;
        a2 = b / j;
        if ((b % a1) == 0) {
          bb = checkPrime(a1);
          if ((bb == true) && (a1 != 1)) {
            long long ss = b;
            long long cnt = 0;
            while ((ss % a1) == 0) {
              cnt++;
              ss /= a1;
            }
            mb.insert(make_pair(a1, cnt));
            ss = a;
            cnt = 0;
            while ((ss % a1) == 0) {
              cnt++;
              ss /= a1;
            }
            ma.insert(make_pair(a1, cnt));
          }
          bb = checkPrime(a2);
          if ((bb == true) && (a2 != 1)) {
            long long ss = b;
            long long cnt = 0;
            while ((ss % a2) == 0) {
              cnt++;
              ss /= a2;
            }
            mb.insert(make_pair(a2, cnt));
            ss = a;
            cnt = 0;
            while ((ss % a2) == 0) {
              cnt++;
              ss /= a2;
            }
            ma.insert(make_pair(a2, cnt));
          }
        }
      }
      long long ans = -1;
      long long kk, pp, diff;
      for (itr = mb.begin(); itr != mb.end(); itr++) {
        diff = abs((itr->second) - ma[(itr->first)]) + 1;
        kk = pow((itr->first), diff);
        pp = a / kk;
        ans = max(ans, pp);
      }
      cout << ans << endl;
    }
  }
}

### Prompt

Develop a solution in Cpp to the problem described below:

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;
bool checkPrime(long long n) {
  long long cnt = 0;
  for (long long i = 2; i <= sqrt(n); i++) {
    if ((n % i) == 0) {
      return false;
    }
  }
  return true;
}
int main() {
  long long t;
  cin >> t;
  while (t--) {
    long long a, b;
    cin >> a >> b;
    if ((a % b) != 0) {
      cout << a << endl;
    } else {
      long long a1, a2;
      bool bb;
      unordered_map<long long, long long> ma;
      unordered_map<long long, long long> mb;
      unordered_map<long long, long long>::iterator itr;
      for (long long j = 1; j <= sqrt(b); j++) {
        a1 = j;
        a2 = b / j;
        if ((b % a1) == 0) {
          bb = checkPrime(a1);
          if ((bb == true) && (a1 != 1)) {
            long long ss = b;
            long long cnt = 0;
            while ((ss % a1) == 0) {
              cnt++;
              ss /= a1;
            }
            mb.insert(make_pair(a1, cnt));
            ss = a;
            cnt = 0;
            while ((ss % a1) == 0) {
              cnt++;
              ss /= a1;
            }
            ma.insert(make_pair(a1, cnt));
          }
          bb = checkPrime(a2);
          if ((bb == true) && (a2 != 1)) {
            long long ss = b;
            long long cnt = 0;
            while ((ss % a2) == 0) {
              cnt++;
              ss /= a2;
            }
            mb.insert(make_pair(a2, cnt));
            ss = a;
            cnt = 0;
            while ((ss % a2) == 0) {
              cnt++;
              ss /= a2;
            }
            ma.insert(make_pair(a2, cnt));
          }
        }
      }
      long long ans = -1;
      long long kk, pp, diff;
      for (itr = mb.begin(); itr != mb.end(); itr++) {
        diff = abs((itr->second) - ma[(itr->first)]) + 1;
        kk = pow((itr->first), diff);
        pp = a / kk;
        ans = max(ans, pp);
      }
      cout << ans << endl;
    }
  }
}

```