#include <bits/stdc++.h>
using namespace std;
template <class T>
inline void tomax(T &x, const T &y) {
  if (x < y) x = y;
}
template <class T>
inline void tomin(T &x, const T &y) {
  if (x > y) x = y;
}
inline long long read() {
  long long w = 0, f = 1;
  char c;
  while (!isdigit(c = getchar())) c == '-' && (f = -1);
  while (isdigit(c)) w = w * 10 + (c ^ 48), c = getchar();
  return w * f;
}
int T, b;
long long a;
int main() {
  int i;
  T = read();
  while (T--) {
    long long ans = 1;
    a = read(), b = read();
    if (a % b != 0) {
      printf("%lld\n", a);
      continue;
    }
    for (i = 2; i * i <= b; i++)
      if (b % i == 0) {
        long long x = a;
        while ((x / i) % b == 0) x /= i;
        x /= i;
        while (b % i == 0) b /= i;
        ans = max(ans, x);
      }
    if (b > 1) {
      long long x = a;
      while ((x / b) % b == 0) x /= b;
      x /= b;
      ans = max(ans, x);
    }
    printf("%lld\n", ans);
  }
  return 0;
}

### Prompt

Please formulate 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;
template <class T>
inline void tomax(T &x, const T &y) {
  if (x < y) x = y;
}
template <class T>
inline void tomin(T &x, const T &y) {
  if (x > y) x = y;
}
inline long long read() {
  long long w = 0, f = 1;
  char c;
  while (!isdigit(c = getchar())) c == '-' && (f = -1);
  while (isdigit(c)) w = w * 10 + (c ^ 48), c = getchar();
  return w * f;
}
int T, b;
long long a;
int main() {
  int i;
  T = read();
  while (T--) {
    long long ans = 1;
    a = read(), b = read();
    if (a % b != 0) {
      printf("%lld\n", a);
      continue;
    }
    for (i = 2; i * i <= b; i++)
      if (b % i == 0) {
        long long x = a;
        while ((x / i) % b == 0) x /= i;
        x /= i;
        while (b % i == 0) b /= i;
        ans = max(ans, x);
      }
    if (b > 1) {
      long long x = a;
      while ((x / b) % b == 0) x /= b;
      x /= b;
      ans = max(ans, x);
    }
    printf("%lld\n", ans);
  }
  return 0;
}

```