#include <bits/stdc++.h>
using namespace std;
long long t, d, ct, q, ans, l, m, r, n, mx, mn, i, j, k, x, y, z, s;
char ch;
string ss = "", str = "";
map<long long, long long> mp;
long long a[200003], b[200003];
long long power(long long a, long long b) {
  long long res = 1;
  for (long long i = 1; i <= b; i++) res *= a;
  return res;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
  cin >> t;
  while (t--) {
    s = k = d = l = mx = m = mn = r = q = i = j = z = x = y = ct = ans = 0;
    cin >> x >> y;
    mx = 1;
    set<long long> st;
    if (y > x || x % y != 0) {
      cout << x << "\n";
      continue;
    }
    m = x;
    d = 1000000000000000000;
    for (i = 2; i * i <= y; i++) {
      if (y % i == 0) {
        s = 0;
        while (y % i == 0) y /= i, s++;
        k = 0;
        while (x % i == 0) x /= i, k++;
        d = min(d, power(i, k - s + 1));
      }
    }
    if (y > 1) {
      k = y;
      s = 0;
      while (x % k == 0) x /= k, s++;
      d = min(d, power(k, s));
    }
    cout << m / d << "\n";
  }
  return 0;
}

### Prompt

Create a solution in Cpp for 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 t, d, ct, q, ans, l, m, r, n, mx, mn, i, j, k, x, y, z, s;
char ch;
string ss = "", str = "";
map<long long, long long> mp;
long long a[200003], b[200003];
long long power(long long a, long long b) {
  long long res = 1;
  for (long long i = 1; i <= b; i++) res *= a;
  return res;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
  cin >> t;
  while (t--) {
    s = k = d = l = mx = m = mn = r = q = i = j = z = x = y = ct = ans = 0;
    cin >> x >> y;
    mx = 1;
    set<long long> st;
    if (y > x || x % y != 0) {
      cout << x << "\n";
      continue;
    }
    m = x;
    d = 1000000000000000000;
    for (i = 2; i * i <= y; i++) {
      if (y % i == 0) {
        s = 0;
        while (y % i == 0) y /= i, s++;
        k = 0;
        while (x % i == 0) x /= i, k++;
        d = min(d, power(i, k - s + 1));
      }
    }
    if (y > 1) {
      k = y;
      s = 0;
      while (x % k == 0) x /= k, s++;
      d = min(d, power(k, s));
    }
    cout << m / d << "\n";
  }
  return 0;
}

```