#include <bits/stdc++.h>
using namespace std;
template <typename T>
void tk_input(vector<T> &arr) {
  for (T &i : arr) cin >> i;
}
const int max_n = 1e5 + 1LL;
int prime[max_n];
void preprocess() {
  prime[1] = 1;
  for (int i = 2; i <= max_n - 1; i++) {
    if (prime[i] == 1) continue;
    int mul = 2;
    while ((mul * i) < max_n) {
      prime[mul * i] = 1;
      mul++;
    }
  }
}
void solve() {
  long long p, q;
  cin >> p >> q;
  if ((p % q) != 0LL) {
    cout << p << '\n';
  } else {
    long long pr = 2LL, ans = 0, curr = p, curr_q = q;
    if ((q % pr) == 0LL) {
      curr = p;
      while ((curr % q) == 0LL) {
        curr /= pr;
      }
      while ((curr_q % pr) == 0LL) {
        curr_q /= pr;
      }
      ans = max(ans, curr);
    }
    for (pr = 3LL; (pr * pr) <= q; pr += 1LL) {
      if ((q % pr) == 0LL) {
        curr = p;
        while ((curr % q) == 0LL) {
          curr /= pr;
        }
        while ((curr_q % pr) == 0LL) {
          curr_q /= pr;
        }
        ans = max(ans, curr);
      }
    }
    if (curr_q != 1LL) {
      curr = p;
      while ((curr % q) == 0LL) {
        curr /= curr_q;
      }
      ans = max(ans, curr);
    }
    cout << ans << '\n';
  }
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  preprocess();
  int _t;
  cin >> _t;
  while (_t--) solve();
}

### 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;
template <typename T>
void tk_input(vector<T> &arr) {
  for (T &i : arr) cin >> i;
}
const int max_n = 1e5 + 1LL;
int prime[max_n];
void preprocess() {
  prime[1] = 1;
  for (int i = 2; i <= max_n - 1; i++) {
    if (prime[i] == 1) continue;
    int mul = 2;
    while ((mul * i) < max_n) {
      prime[mul * i] = 1;
      mul++;
    }
  }
}
void solve() {
  long long p, q;
  cin >> p >> q;
  if ((p % q) != 0LL) {
    cout << p << '\n';
  } else {
    long long pr = 2LL, ans = 0, curr = p, curr_q = q;
    if ((q % pr) == 0LL) {
      curr = p;
      while ((curr % q) == 0LL) {
        curr /= pr;
      }
      while ((curr_q % pr) == 0LL) {
        curr_q /= pr;
      }
      ans = max(ans, curr);
    }
    for (pr = 3LL; (pr * pr) <= q; pr += 1LL) {
      if ((q % pr) == 0LL) {
        curr = p;
        while ((curr % q) == 0LL) {
          curr /= pr;
        }
        while ((curr_q % pr) == 0LL) {
          curr_q /= pr;
        }
        ans = max(ans, curr);
      }
    }
    if (curr_q != 1LL) {
      curr = p;
      while ((curr % q) == 0LL) {
        curr /= curr_q;
      }
      ans = max(ans, curr);
    }
    cout << ans << '\n';
  }
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  preprocess();
  int _t;
  cin >> _t;
  while (_t--) solve();
}

```