#include <bits/stdc++.h>
using namespace std;
struct keycompare {
  bool operator()(const pair<long long, long long>& v, const long long& k) {
    return (v.first < k);
  }
  bool operator()(const long long& k, const pair<long long, long long>& v) {
    return (k < v.first);
  }
};
long long mod1 = 998244353, mod2 = 1000000007, limit = 9223372036854775807;
long double pi = 3.1415926535897932;
long long modpow(long long x, long long n, long long m) {
  if (x > m) {
    x %= m;
  }
  if (n == 0) return 1 % m;
  long long u = modpow(x, n / 2, m);
  u = (u * u) % m;
  if (n % 2 == 1) u = (u * x) % m;
  return u;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
bool isprime(long long n) {
  if (n == 2) return true;
  for (long long i = 2; i * i <= n; i++) {
    if (n % i == 0) return false;
  }
  return true;
}
long long power(long long x, long long n) {
  long long x_n = 1;
  for (long long i = 0; i < n; i++) {
    x_n *= x;
  }
  return x_n;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t;
  cin >> t;
  while (t--) {
    long long p, q;
    cin >> p >> q;
    vector<long long> v;
    for (long long i = 2; i * i <= q; i++) {
      if (q % i == 0) {
        v.push_back(i);
        v.push_back(q / i);
      }
    }
    v.push_back(q);
    long long ans = 1;
    for (auto it : v) {
      long long temp = 0;
      long long tempp = p;
      while (tempp % q == 0) {
        tempp /= it;
      }
      ans = max(ans, tempp);
    }
    cout << ans << endl;
  }
}

### Prompt

Please create a solution in Cpp 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;
struct keycompare {
  bool operator()(const pair<long long, long long>& v, const long long& k) {
    return (v.first < k);
  }
  bool operator()(const long long& k, const pair<long long, long long>& v) {
    return (k < v.first);
  }
};
long long mod1 = 998244353, mod2 = 1000000007, limit = 9223372036854775807;
long double pi = 3.1415926535897932;
long long modpow(long long x, long long n, long long m) {
  if (x > m) {
    x %= m;
  }
  if (n == 0) return 1 % m;
  long long u = modpow(x, n / 2, m);
  u = (u * u) % m;
  if (n % 2 == 1) u = (u * x) % m;
  return u;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
bool isprime(long long n) {
  if (n == 2) return true;
  for (long long i = 2; i * i <= n; i++) {
    if (n % i == 0) return false;
  }
  return true;
}
long long power(long long x, long long n) {
  long long x_n = 1;
  for (long long i = 0; i < n; i++) {
    x_n *= x;
  }
  return x_n;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t;
  cin >> t;
  while (t--) {
    long long p, q;
    cin >> p >> q;
    vector<long long> v;
    for (long long i = 2; i * i <= q; i++) {
      if (q % i == 0) {
        v.push_back(i);
        v.push_back(q / i);
      }
    }
    v.push_back(q);
    long long ans = 1;
    for (auto it : v) {
      long long temp = 0;
      long long tempp = p;
      while (tempp % q == 0) {
        tempp /= it;
      }
      ans = max(ans, tempp);
    }
    cout << ans << endl;
  }
}

```