#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int MX = 1e9;
bool is_prime(ll x) {
  if (x < 2) return false;
  if (x == 2) return true;
  for (ll i = 2; i * i <= x; ++i) {
    if (x % i == 0) return false;
  }
  return true;
}
vector<ll> f(ll x) {
  vector<ll> res;
  for (ll i = 1; i * i <= x; ++i) {
    if (x % i == 0) {
      res.push_back(i);
      ll j = x / i;
      if (x % j == 0 && j != i) {
        res.push_back(j);
      }
    }
  }
  return res;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int t;
  cin >> t;
  while (t--) {
    ll p, q;
    cin >> p >> q;
    if (p < q || p % q != 0) {
      cout << p << '\n';
      continue;
    }
    if (p <= MX) {
      auto d = f(p);
      ll res = 1;
      for (ll x : d) {
        if (x % q != 0) res = max(res, x);
      }
      cout << res << '\n';
    } else {
      ll res = p;
      while (res % q == 0) res /= q;
      for (ll i = 1; i <= 1e5; ++i) {
        if (p % i == 0) {
          if (i % q != 0) {
            res = max(i, res);
          }
          ll x = p;
          while (i > 1 && x > 0) {
            if (p % x == 0 && x % q != 0) {
              res = max(x, res);
            }
            x /= i;
          }
        }
      }
      cout << res << '\n';
    }
  }
}

### Prompt

Please provide a Cpp coded solution 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;
using ll = long long;
const int MX = 1e9;
bool is_prime(ll x) {
  if (x < 2) return false;
  if (x == 2) return true;
  for (ll i = 2; i * i <= x; ++i) {
    if (x % i == 0) return false;
  }
  return true;
}
vector<ll> f(ll x) {
  vector<ll> res;
  for (ll i = 1; i * i <= x; ++i) {
    if (x % i == 0) {
      res.push_back(i);
      ll j = x / i;
      if (x % j == 0 && j != i) {
        res.push_back(j);
      }
    }
  }
  return res;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int t;
  cin >> t;
  while (t--) {
    ll p, q;
    cin >> p >> q;
    if (p < q || p % q != 0) {
      cout << p << '\n';
      continue;
    }
    if (p <= MX) {
      auto d = f(p);
      ll res = 1;
      for (ll x : d) {
        if (x % q != 0) res = max(res, x);
      }
      cout << res << '\n';
    } else {
      ll res = p;
      while (res % q == 0) res /= q;
      for (ll i = 1; i <= 1e5; ++i) {
        if (p % i == 0) {
          if (i % q != 0) {
            res = max(i, res);
          }
          ll x = p;
          while (i > 1 && x > 0) {
            if (p % x == 0 && x % q != 0) {
              res = max(x, res);
            }
            x /= i;
          }
        }
      }
      cout << res << '\n';
    }
  }
}

```