#include <bits/stdc++.h>
using namespace std;
long long M = 1000000007;
void fast() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
long long power(long long base, long long n) {
  if (n == 0) return 1;
  if (n == 1) return base;
  long long halfn = power(base, n / 2);
  if (n % 2 == 0)
    return (halfn * halfn) % M;
  else
    return (((halfn * halfn) % M) * base) % M;
}
void solve() {
  long long p, q;
  cin >> p >> q;
  if (p % q != 0) {
    cout << p << '\n';
    return;
  }
  vector<long long> v;
  long long z = sqrt(q), i;
  for (i = 2; i <= z; i++) {
    if (q % i == 0) {
      v.push_back(i);
      if (i != (q / i)) v.push_back(q / i);
    }
  }
  sort(v.begin(), v.end());
  v.push_back(q);
  long long x = p / q;
  long long ans = 1;
  for (long long i = v.size() - 1; i >= 0; i--) {
    long long an = 1, xx = x;
    while (xx % v[i] == 0) xx /= v[i];
    an = xx * (q / v[i]);
    if (an % q) ans = max(ans, an);
  }
  cout << ans << '\n';
}
int main() {
  fast();
  long long tt;
  cin >> tt;
  while (tt--) {
    solve();
  }
  return 0;
}

### 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;
long long M = 1000000007;
void fast() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
long long power(long long base, long long n) {
  if (n == 0) return 1;
  if (n == 1) return base;
  long long halfn = power(base, n / 2);
  if (n % 2 == 0)
    return (halfn * halfn) % M;
  else
    return (((halfn * halfn) % M) * base) % M;
}
void solve() {
  long long p, q;
  cin >> p >> q;
  if (p % q != 0) {
    cout << p << '\n';
    return;
  }
  vector<long long> v;
  long long z = sqrt(q), i;
  for (i = 2; i <= z; i++) {
    if (q % i == 0) {
      v.push_back(i);
      if (i != (q / i)) v.push_back(q / i);
    }
  }
  sort(v.begin(), v.end());
  v.push_back(q);
  long long x = p / q;
  long long ans = 1;
  for (long long i = v.size() - 1; i >= 0; i--) {
    long long an = 1, xx = x;
    while (xx % v[i] == 0) xx /= v[i];
    an = xx * (q / v[i]);
    if (an % q) ans = max(ans, an);
  }
  cout << ans << '\n';
}
int main() {
  fast();
  long long tt;
  cin >> tt;
  while (tt--) {
    solve();
  }
  return 0;
}

```