#include <bits/stdc++.h>
using namespace std;
void onami(long long n, long long b, long long a) {
  map<long long, long long> m;
  map<long long, long long>::iterator it;
  while (n % 2 == 0) {
    m[2]++;
    n = n / 2;
  }
  for (long long i = 3; i <= sqrt(n); i = i + 2) {
    while (n % i == 0) {
      m[i]++;
      n = n / i;
    }
  }
  if (n > 2) {
    m[n]++;
  }
  long long xx, y;
  xx = 1;
  y = 1;
  for (it = m.begin(); it != m.end(); it++) {
    long long nnn = a;
    while (nnn % b == 0) {
      nnn = nnn / (*it).first;
    }
    xx = max(nnn, xx);
  }
  n = b;
  vector<long long> v;
  for (long long i = 1; i <= sqrt(n); i++) {
    if (n % i == 0) {
      if (n / i == i) {
        v.push_back(i);
      } else {
        v.push_back(i);
        v.push_back(n / i);
      }
    }
  }
  sort(v.rbegin(), v.rend());
  y = v[1];
  cout << max(xx, y);
}
int main() {
  int Tt;
  cin >> Tt;
  for (int i = 0; i < Tt; i++) {
    long long a, b, c;
    cin >> a;
    cin >> b;
    if (b <= a && a % b == 0) {
      onami(b, b, a);
      printf("\n");
    } else {
      cout << a;
      printf("\n");
    }
  }
}

### Prompt

In CPP, your task is to solve 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;
void onami(long long n, long long b, long long a) {
  map<long long, long long> m;
  map<long long, long long>::iterator it;
  while (n % 2 == 0) {
    m[2]++;
    n = n / 2;
  }
  for (long long i = 3; i <= sqrt(n); i = i + 2) {
    while (n % i == 0) {
      m[i]++;
      n = n / i;
    }
  }
  if (n > 2) {
    m[n]++;
  }
  long long xx, y;
  xx = 1;
  y = 1;
  for (it = m.begin(); it != m.end(); it++) {
    long long nnn = a;
    while (nnn % b == 0) {
      nnn = nnn / (*it).first;
    }
    xx = max(nnn, xx);
  }
  n = b;
  vector<long long> v;
  for (long long i = 1; i <= sqrt(n); i++) {
    if (n % i == 0) {
      if (n / i == i) {
        v.push_back(i);
      } else {
        v.push_back(i);
        v.push_back(n / i);
      }
    }
  }
  sort(v.rbegin(), v.rend());
  y = v[1];
  cout << max(xx, y);
}
int main() {
  int Tt;
  cin >> Tt;
  for (int i = 0; i < Tt; i++) {
    long long a, b, c;
    cin >> a;
    cin >> b;
    if (b <= a && a % b == 0) {
      onami(b, b, a);
      printf("\n");
    } else {
      cout << a;
      printf("\n");
    }
  }
}

```