#include <bits/stdc++.h>
using namespace std;
int cases(1), t;
template <typename T>
void binary(T n) {
  stack<int> st;
  while (n > 0) {
    st.push(n % 2);
    n /= 2;
  }
  while (!st.empty()) {
    cout << st.top();
    st.pop();
  }
  cout << '\n';
}
vector<long long> factorize(long long x) {
  vector<long long> res;
  if (x % 2 == 0) {
    res.push_back(2);
  }
  while (x % 2 == 0) {
    x /= 2;
  }
  for (long long i = 3; i * i <= x; i += 2) {
    if (x % i == 0) {
      res.push_back(i);
      while (x % i == 0) {
        x /= i;
      }
    }
  }
  if (x > 2) res.push_back(x);
  return res;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  for (cin >> t; cases <= t; ++cases) {
    long long first, second;
    cin >> first >> second;
    long long ans = 0;
    if (first % second) {
      cout << first << '\n';
      continue;
    }
    vector<long long> fac = factorize(second);
    for (long long i = 0; i < fac.size(); i++) {
      long long res = first;
      while (res % second == 0) {
        res /= fac[i];
      }
      ans = max(res, ans);
    }
    cout << ans << '\n';
  }
  return 0;
}

### 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;
int cases(1), t;
template <typename T>
void binary(T n) {
  stack<int> st;
  while (n > 0) {
    st.push(n % 2);
    n /= 2;
  }
  while (!st.empty()) {
    cout << st.top();
    st.pop();
  }
  cout << '\n';
}
vector<long long> factorize(long long x) {
  vector<long long> res;
  if (x % 2 == 0) {
    res.push_back(2);
  }
  while (x % 2 == 0) {
    x /= 2;
  }
  for (long long i = 3; i * i <= x; i += 2) {
    if (x % i == 0) {
      res.push_back(i);
      while (x % i == 0) {
        x /= i;
      }
    }
  }
  if (x > 2) res.push_back(x);
  return res;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  for (cin >> t; cases <= t; ++cases) {
    long long first, second;
    cin >> first >> second;
    long long ans = 0;
    if (first % second) {
      cout << first << '\n';
      continue;
    }
    vector<long long> fac = factorize(second);
    for (long long i = 0; i < fac.size(); i++) {
      long long res = first;
      while (res % second == 0) {
        res /= fac[i];
      }
      ans = max(res, ans);
    }
    cout << ans << '\n';
  }
  return 0;
}

```