#include <bits/stdc++.h>
using namespace std;
const int lim = 100005;
bool np[lim];
int p[lim];
int tot;
int k[lim];
int yinzi[lim];
int phi[lim];
void getphi() {
  np[1] = phi[1] = 1;
  for (int i = 2; i <= lim; ++i) {
    if (!np[i]) p[++tot] = i, phi[i] = i - 1;
    for (int j = 1; j <= tot && i * p[j] <= lim; ++j) {
      np[i * p[j]] = 1;
      if (i % p[j] == 0) {
        phi[i * p[j]] = phi[i] * p[j];
        break;
      } else
        phi[i * p[j]] = phi[i] * (p[j] - 1);
    }
  }
}
int main() {
  int T;
  cin >> T;
  getphi();
  while (T--) {
    long long a, b;
    long long ans = 0;
    int cnt = 0;
    cin >> a >> b;
    if (a < b)
      cout << a << endl;
    else if (a % b != 0)
      cout << a << endl;
    else {
      for (int i = 1; i <= tot; ++i) {
        if (b % p[i] != 0) continue;
        long long res = 1;
        while (b % p[i] == 0) {
          b /= p[i];
          res *= p[i];
        }
        res /= p[i];
        long long a_res = a;
        while (a_res % p[i] == 0) {
          a_res /= p[i];
        }
        ans = max(ans, res * a_res);
      }
      if (b != 1) {
        long long res = 1;
        long long a_res = a;
        while (a_res % b == 0) {
          a_res /= b;
        }
        ans = max(ans, res * a_res);
      }
      cout << ans << endl;
    }
  }
  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;
const int lim = 100005;
bool np[lim];
int p[lim];
int tot;
int k[lim];
int yinzi[lim];
int phi[lim];
void getphi() {
  np[1] = phi[1] = 1;
  for (int i = 2; i <= lim; ++i) {
    if (!np[i]) p[++tot] = i, phi[i] = i - 1;
    for (int j = 1; j <= tot && i * p[j] <= lim; ++j) {
      np[i * p[j]] = 1;
      if (i % p[j] == 0) {
        phi[i * p[j]] = phi[i] * p[j];
        break;
      } else
        phi[i * p[j]] = phi[i] * (p[j] - 1);
    }
  }
}
int main() {
  int T;
  cin >> T;
  getphi();
  while (T--) {
    long long a, b;
    long long ans = 0;
    int cnt = 0;
    cin >> a >> b;
    if (a < b)
      cout << a << endl;
    else if (a % b != 0)
      cout << a << endl;
    else {
      for (int i = 1; i <= tot; ++i) {
        if (b % p[i] != 0) continue;
        long long res = 1;
        while (b % p[i] == 0) {
          b /= p[i];
          res *= p[i];
        }
        res /= p[i];
        long long a_res = a;
        while (a_res % p[i] == 0) {
          a_res /= p[i];
        }
        ans = max(ans, res * a_res);
      }
      if (b != 1) {
        long long res = 1;
        long long a_res = a;
        while (a_res % b == 0) {
          a_res /= b;
        }
        ans = max(ans, res * a_res);
      }
      cout << ans << endl;
    }
  }
  return 0;
}

```