#include <bits/stdc++.h>
using namespace std;
template <class T>
bool tmin(T& a, const T& b) {
  return b < a ? a = b, 1 : 0;
}
template <class T>
bool tmax(T& a, const T& b) {
  return a < b ? a = b, 1 : 0;
}
void setIO() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
}
long long P, Q;
long long p, q, r;
bool upd(long long pr) {
  int fp = 0, fq = 0;
  while (q > 1 && q % pr == 0) {
    ++fq;
    q /= pr;
  }
  if (fq == 0) return false;
  while (p > 1 && p % pr == 0) {
    ++fp;
    p /= pr;
  }
  if (fq > fp) return true;
  long long rr = 1;
  for (int _n(fp - fq + 1), i(0); i < _n; i++) rr *= pr;
  r = min(r, rr);
  return false;
}
long long solve() {
  if (P < Q) return P;
  r = P;
  p = P;
  q = Q;
  if (upd(2)) return P;
  for (long long pr = 3; pr * pr <= Q; pr += 2)
    if (upd(pr)) return P;
  if (q > 1 && upd(q)) return P;
  return P / r;
}
int main() {
  setIO();
  int t;
  cin >> t;
  while (t--) {
    cin >> P >> Q;
    cout << solve() << "\n";
  }
  return 0;
}

### Prompt

Generate a Cpp solution to 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;
template <class T>
bool tmin(T& a, const T& b) {
  return b < a ? a = b, 1 : 0;
}
template <class T>
bool tmax(T& a, const T& b) {
  return a < b ? a = b, 1 : 0;
}
void setIO() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
}
long long P, Q;
long long p, q, r;
bool upd(long long pr) {
  int fp = 0, fq = 0;
  while (q > 1 && q % pr == 0) {
    ++fq;
    q /= pr;
  }
  if (fq == 0) return false;
  while (p > 1 && p % pr == 0) {
    ++fp;
    p /= pr;
  }
  if (fq > fp) return true;
  long long rr = 1;
  for (int _n(fp - fq + 1), i(0); i < _n; i++) rr *= pr;
  r = min(r, rr);
  return false;
}
long long solve() {
  if (P < Q) return P;
  r = P;
  p = P;
  q = Q;
  if (upd(2)) return P;
  for (long long pr = 3; pr * pr <= Q; pr += 2)
    if (upd(pr)) return P;
  if (q > 1 && upd(q)) return P;
  return P / r;
}
int main() {
  setIO();
  int t;
  cin >> t;
  while (t--) {
    cin >> P >> Q;
    cout << solve() << "\n";
  }
  return 0;
}

```