#include <bits/stdc++.h>
using namespace std;
long long int pow(long long int a, long long int b, long long int m) {
  long long int ans = 1;
  while (b) {
    if (b & 1) {
      ans = (ans * a) % m;
      b--;
    }
    a = (a * a) % m;
    b /= 2;
  }
  return ans;
}
long long int inverse(long long int a, long long int m) {
  return pow(a, m - 2, m);
}
long long int ncr(long long int n, long long int r) {
  if (r > n / 2) r = n - r;
  long long int temp = r, ans = 1, ans2 = 1;
  while (temp != 0) {
    ans = (ans * n) % 1000000007;
    n--;
    temp--;
  }
  while (r != 0) {
    ans2 = (ans2 * r) % 1000000007;
    r--;
  }
  ans = (ans * (inverse(ans2, 1000000007))) % 1000000007;
  return ans;
}
int main() {
  long long int t;
  cin >> t;
  while (t--) {
    long long int a, b, flag = 0, temp = 0, mx = 0;
    cin >> a >> b;
    vector<long long int> v;
    temp = b;
    for (long long int i = 2; i <= sqrt(b); i++) {
      if (temp % i == 0) {
        v.push_back(i);
        if (i != temp / i) v.push_back(b / i);
        while (temp % i == 0) temp /= i;
      }
    }
    if (temp != 1) v.push_back(temp);
    for (long long int i = 0; i < v.size(); i++) {
      temp = a;
      while (temp % v[i] == 0) {
        temp /= v[i];
        if (temp % b != 0) {
          flag = 1;
          mx = max(mx, temp);
        }
      }
    }
    if (a % b != 0)
      cout << a << endl;
    else if (flag == 1)
      cout << mx << endl;
    else
      cout << "1" << endl;
  }
  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;
long long int pow(long long int a, long long int b, long long int m) {
  long long int ans = 1;
  while (b) {
    if (b & 1) {
      ans = (ans * a) % m;
      b--;
    }
    a = (a * a) % m;
    b /= 2;
  }
  return ans;
}
long long int inverse(long long int a, long long int m) {
  return pow(a, m - 2, m);
}
long long int ncr(long long int n, long long int r) {
  if (r > n / 2) r = n - r;
  long long int temp = r, ans = 1, ans2 = 1;
  while (temp != 0) {
    ans = (ans * n) % 1000000007;
    n--;
    temp--;
  }
  while (r != 0) {
    ans2 = (ans2 * r) % 1000000007;
    r--;
  }
  ans = (ans * (inverse(ans2, 1000000007))) % 1000000007;
  return ans;
}
int main() {
  long long int t;
  cin >> t;
  while (t--) {
    long long int a, b, flag = 0, temp = 0, mx = 0;
    cin >> a >> b;
    vector<long long int> v;
    temp = b;
    for (long long int i = 2; i <= sqrt(b); i++) {
      if (temp % i == 0) {
        v.push_back(i);
        if (i != temp / i) v.push_back(b / i);
        while (temp % i == 0) temp /= i;
      }
    }
    if (temp != 1) v.push_back(temp);
    for (long long int i = 0; i < v.size(); i++) {
      temp = a;
      while (temp % v[i] == 0) {
        temp /= v[i];
        if (temp % b != 0) {
          flag = 1;
          mx = max(mx, temp);
        }
      }
    }
    if (a % b != 0)
      cout << a << endl;
    else if (flag == 1)
      cout << mx << endl;
    else
      cout << "1" << endl;
  }
  return 0;
}

```