#include <bits/stdc++.h>
using namespace std;
void SieveOfEratosthenes(long long n) {
  bool prime[n + 1];
  memset(prime, true, sizeof(prime));
  for (long long p = 2; p * p <= n; p++) {
    if (prime[p] == true) {
      for (long long i = p * p; i <= n; i += p) prime[i] = false;
    }
  }
  for (long long p = 2; p <= n; p++)
    if (prime[p]) cout << p << " ";
}
bool isPrime(long long n) {
  if (n <= 1) return false;
  for (long long i = 2; i * i <= n; i++)
    if (n % i == 0) return false;
  return true;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long a;
  cin >> a;
  long long o = pow(10, 6);
  bool prime[o + 1];
  memset(prime, true, sizeof(prime));
  for (long long p = 2; p * p <= o; p++) {
    if (prime[p] == true) {
      for (long long i = p * p; i <= o; i += p) prime[i] = false;
    }
  }
  while (a--) {
    long long n, m;
    cin >> n >> m;
    vector<long long> v1;
    long long z = m;
    for (long long i = 2; i * i <= z; i++) {
      long long count = 0;
      if (prime[i]) {
        while (z % i == 0) {
          z = z / i;
          count++;
        }
        v1.push_back(i);
      }
    }
    if (z > 1) v1.push_back(z);
    long long ans = 1;
    if (n % m != 0) ans = n;
    for (long long i = 0; i < v1.size(); i++) {
      long long k = n;
      while (k % v1[i] == 0) {
        k = k / v1[i];
        if (k % m != 0) {
          ans = max(k, ans);
        }
      }
    }
    cout << ans << "\n";
  }
}

### 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;
void SieveOfEratosthenes(long long n) {
  bool prime[n + 1];
  memset(prime, true, sizeof(prime));
  for (long long p = 2; p * p <= n; p++) {
    if (prime[p] == true) {
      for (long long i = p * p; i <= n; i += p) prime[i] = false;
    }
  }
  for (long long p = 2; p <= n; p++)
    if (prime[p]) cout << p << " ";
}
bool isPrime(long long n) {
  if (n <= 1) return false;
  for (long long i = 2; i * i <= n; i++)
    if (n % i == 0) return false;
  return true;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long a;
  cin >> a;
  long long o = pow(10, 6);
  bool prime[o + 1];
  memset(prime, true, sizeof(prime));
  for (long long p = 2; p * p <= o; p++) {
    if (prime[p] == true) {
      for (long long i = p * p; i <= o; i += p) prime[i] = false;
    }
  }
  while (a--) {
    long long n, m;
    cin >> n >> m;
    vector<long long> v1;
    long long z = m;
    for (long long i = 2; i * i <= z; i++) {
      long long count = 0;
      if (prime[i]) {
        while (z % i == 0) {
          z = z / i;
          count++;
        }
        v1.push_back(i);
      }
    }
    if (z > 1) v1.push_back(z);
    long long ans = 1;
    if (n % m != 0) ans = n;
    for (long long i = 0; i < v1.size(); i++) {
      long long k = n;
      while (k % v1[i] == 0) {
        k = k / v1[i];
        if (k % m != 0) {
          ans = max(k, ans);
        }
      }
    }
    cout << ans << "\n";
  }
}

```