#include <bits/stdc++.h>
using namespace std;
bool not_prime[700000];
vector<long long> primes;
void seive() {
  long long i, j, lim = sqrt(700000);
  for (i = 3; i <= lim; i += 2) {
    if (not_prime[i] == 0) {
      for (j = i * i; j <= 700000; j += i + i) {
        not_prime[j] = 1;
      }
    }
  }
  primes.push_back(2);
  for (i = 3; i <= 700000; i += 2) {
    if (not_prime[i] == 0) {
      primes.push_back(i);
    }
  }
}
int main() {
  long long t, i, n, x, y, m, a, b, c, d, ans;
  seive();
  cin >> t;
  while (t--) {
    cin >> a >> b;
    if (a % b != 0) {
      ans = a;
    } else {
      vector<long long> primefact;
      ans = LLONG_MIN;
      y = b;
      x = sqrt(y);
      for (i = 0; primes[i] <= x; i++) {
        if (y % primes[i] == 0) {
          while (y % primes[i] == 0) {
            y /= primes[i];
          }
          primefact.push_back(primes[i]);
          x = sqrt(y);
        }
      }
      if (y > 1) {
        primefact.push_back(y);
      }
      for (i = 0; i < primefact.size(); i++) {
        x = a;
        if (x % primefact[i] == 0) {
          while (x % primefact[i] == 0) {
            x /= primefact[i];
            if (x % b != 0) {
              ans = max(ans, x);
            }
          }
        }
      }
    }
    cout << ans << endl;
  }
  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;
bool not_prime[700000];
vector<long long> primes;
void seive() {
  long long i, j, lim = sqrt(700000);
  for (i = 3; i <= lim; i += 2) {
    if (not_prime[i] == 0) {
      for (j = i * i; j <= 700000; j += i + i) {
        not_prime[j] = 1;
      }
    }
  }
  primes.push_back(2);
  for (i = 3; i <= 700000; i += 2) {
    if (not_prime[i] == 0) {
      primes.push_back(i);
    }
  }
}
int main() {
  long long t, i, n, x, y, m, a, b, c, d, ans;
  seive();
  cin >> t;
  while (t--) {
    cin >> a >> b;
    if (a % b != 0) {
      ans = a;
    } else {
      vector<long long> primefact;
      ans = LLONG_MIN;
      y = b;
      x = sqrt(y);
      for (i = 0; primes[i] <= x; i++) {
        if (y % primes[i] == 0) {
          while (y % primes[i] == 0) {
            y /= primes[i];
          }
          primefact.push_back(primes[i]);
          x = sqrt(y);
        }
      }
      if (y > 1) {
        primefact.push_back(y);
      }
      for (i = 0; i < primefact.size(); i++) {
        x = a;
        if (x % primefact[i] == 0) {
          while (x % primefact[i] == 0) {
            x /= primefact[i];
            if (x % b != 0) {
              ans = max(ans, x);
            }
          }
        }
      }
    }
    cout << ans << endl;
  }
  return 0;
}

```