#include <bits/stdc++.h>
using namespace std;
long long signed pow_(long long signed a, long long signed e,
                      long long signed m = 1000000007) {
  long long signed r = 1;
  while (e) {
    if (e & 1) r = r * a % m;
    a = a * a % m;
    e >>= 1;
  }
  return r;
}
long long inv(long long signed a) {
  return pow_(a, 1000000007 - 2, 1000000007);
}
long long add(long long signed a, long long signed b) {
  return (a + b + 1000000007) % 1000000007;
}
long long mul(long long signed a, long long signed b) {
  return (1000000007 + a * b % 1000000007) % 1000000007;
}
long long di(long long signed a, long long signed b) { return mul(a, inv(b)); }
long long signed ceil(long long signed a, long long signed b) {
  return a / b + (a % b != 0);
}
void print(vector<long long> a) {
  for (long long i = 0; i < a.size(); i++) cout << a[i] << " ";
  cout << '\n';
}
void print(long long signed a, long long signed b, char k = '\n') {
  cout << a << " " << b << k;
}
void print(vector<long long> *a, long long n) {
  for (long long i = 0; i < n; i++) print(a[i]);
}
void print(long long *a, long long n) {
  for (long long i = 0; i < n; i++) cout << a[i] << " ";
  cout << '\n';
}
long long arr[1000005];
long long signed A, B;
bool v[1000005];
long long sp[1000005];
vector<long long> PM;
void Sieve() {
  PM.push_back(2);
  for (long long i = 2; i < 1000005; i += 2) sp[i] = 2;
  for (long long signed i = 3; i < 1000005; i += 2) {
    if (!v[i]) {
      PM.push_back(i);
      sp[i] = i;
      for (long long signed j = i; (j * i) < 1000005; j += 2) {
        if (!v[j * i]) v[j * i] = true, sp[j * i] = i;
      }
    }
  }
}
long long solve() {
  cin >> A >> B;
  bool f = false;
  long long signed R = 1;
  for (auto i : PM) {
    if (B % i == 0) {
      long long signed X = A;
      while (X % B == 0) X /= i;
      R = max(R, X);
    }
  }
  while (A % B == 0) {
    A /= B;
  }
  cout << max(R, A) << '\n';
  return 0;
}
signed main() {
  Sieve();
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  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;
long long signed pow_(long long signed a, long long signed e,
                      long long signed m = 1000000007) {
  long long signed r = 1;
  while (e) {
    if (e & 1) r = r * a % m;
    a = a * a % m;
    e >>= 1;
  }
  return r;
}
long long inv(long long signed a) {
  return pow_(a, 1000000007 - 2, 1000000007);
}
long long add(long long signed a, long long signed b) {
  return (a + b + 1000000007) % 1000000007;
}
long long mul(long long signed a, long long signed b) {
  return (1000000007 + a * b % 1000000007) % 1000000007;
}
long long di(long long signed a, long long signed b) { return mul(a, inv(b)); }
long long signed ceil(long long signed a, long long signed b) {
  return a / b + (a % b != 0);
}
void print(vector<long long> a) {
  for (long long i = 0; i < a.size(); i++) cout << a[i] << " ";
  cout << '\n';
}
void print(long long signed a, long long signed b, char k = '\n') {
  cout << a << " " << b << k;
}
void print(vector<long long> *a, long long n) {
  for (long long i = 0; i < n; i++) print(a[i]);
}
void print(long long *a, long long n) {
  for (long long i = 0; i < n; i++) cout << a[i] << " ";
  cout << '\n';
}
long long arr[1000005];
long long signed A, B;
bool v[1000005];
long long sp[1000005];
vector<long long> PM;
void Sieve() {
  PM.push_back(2);
  for (long long i = 2; i < 1000005; i += 2) sp[i] = 2;
  for (long long signed i = 3; i < 1000005; i += 2) {
    if (!v[i]) {
      PM.push_back(i);
      sp[i] = i;
      for (long long signed j = i; (j * i) < 1000005; j += 2) {
        if (!v[j * i]) v[j * i] = true, sp[j * i] = i;
      }
    }
  }
}
long long solve() {
  cin >> A >> B;
  bool f = false;
  long long signed R = 1;
  for (auto i : PM) {
    if (B % i == 0) {
      long long signed X = A;
      while (X % B == 0) X /= i;
      R = max(R, X);
    }
  }
  while (A % B == 0) {
    A /= B;
  }
  cout << max(R, A) << '\n';
  return 0;
}
signed main() {
  Sieve();
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

```