#include <bits/stdc++.h>
using namespace std;
const long long maxn = 1e7;
long long is_prime[maxn], prime[maxn];
int aishishai(int ub) {
  int p = 0;
  memset(is_prime, 0, sizeof(is_prime));
  for (int i = 2; i <= ub; ++i) {
    if (is_prime[i]) continue;
    prime[p++] = i;
    for (int j = i; j < ub; j += i) is_prime[j] = 1;
  }
  return p;
}
int main() {
  int t;
  cin >> t;
  int len = aishishai(maxn - 1);
  while (t--) {
    long long p, q;
    cin >> p >> q;
    long long qq = q;
    if (p % q != 0) {
      cout << p << '\n';
      continue;
    }
    int val = 1;
    long long Max = 0;
    for (int i = 0; prime[i] <= q && i < len; ++i) {
      if (q % prime[i] != 0) continue;
      long long pp = p;
      val = 1;
      while (pp % prime[i] == 0) pp /= prime[i];
      while (q % prime[i] == 0) q /= prime[i], val *= prime[i];
      Max = max(Max, pp * val / prime[i]);
    }
    while (p % qq == 0 && p) {
      p /= qq;
    }
    if (Max == 0)
      cout << p << '\n';
    else
      cout << Max << '\n';
  }
}

### Prompt

Create a solution in CPP for 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;
const long long maxn = 1e7;
long long is_prime[maxn], prime[maxn];
int aishishai(int ub) {
  int p = 0;
  memset(is_prime, 0, sizeof(is_prime));
  for (int i = 2; i <= ub; ++i) {
    if (is_prime[i]) continue;
    prime[p++] = i;
    for (int j = i; j < ub; j += i) is_prime[j] = 1;
  }
  return p;
}
int main() {
  int t;
  cin >> t;
  int len = aishishai(maxn - 1);
  while (t--) {
    long long p, q;
    cin >> p >> q;
    long long qq = q;
    if (p % q != 0) {
      cout << p << '\n';
      continue;
    }
    int val = 1;
    long long Max = 0;
    for (int i = 0; prime[i] <= q && i < len; ++i) {
      if (q % prime[i] != 0) continue;
      long long pp = p;
      val = 1;
      while (pp % prime[i] == 0) pp /= prime[i];
      while (q % prime[i] == 0) q /= prime[i], val *= prime[i];
      Max = max(Max, pp * val / prime[i]);
    }
    while (p % qq == 0 && p) {
      p /= qq;
    }
    if (Max == 0)
      cout << p << '\n';
    else
      cout << Max << '\n';
  }
}

```