#include <bits/stdc++.h>
using namespace std;
vector<long long> prime;
vector<bool> vis(1e5 + 1000, true);
void sieve(long long n) {
  prime.push_back(2);
  for (long long i = 4; i <= n; i += 2) {
    vis[i] = false;
  }
  for (long long i = 3; i <= n; i += 2) {
    if (vis[i]) {
      prime.push_back(i);
      if (i > sqrt(n)) {
        continue;
      }
      for (long long j = i * i; j <= n; j += 2 * i) {
        vis[j] = false;
      }
    }
  }
}
int main() {
  sieve(100100);
  int t;
  cin >> t;
  for (int tc = 1; tc <= t; tc++) {
    long long p, q;
    cin >> p >> q;
    if (p % q != 0) {
      cout << p << '\n';
      continue;
    }
    vector<long long> factor, power;
    for (int i = 0; prime[i] <= sqrt(q); i++) {
      if (q % prime[i] == 0) {
        factor.push_back(prime[i]);
        long long cur = 0;
        while (q % prime[i] == 0) {
          cur++;
          q /= prime[i];
        }
        power.push_back(cur);
      }
    }
    if (q != 1) {
      factor.push_back(q);
      power.push_back(1);
    }
    long long ans = 0;
    for (int i = 0; i < factor.size(); i++) {
      long long r = factor[i];
      long long cur = p;
      while (cur % r == 0) {
        cur /= r;
      }
      for (long long j = 1; j < power[i]; j++) {
        cur *= r;
      }
      ans = max(ans, cur);
    }
    cout << ans << '\n';
  }
}

### 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;
vector<long long> prime;
vector<bool> vis(1e5 + 1000, true);
void sieve(long long n) {
  prime.push_back(2);
  for (long long i = 4; i <= n; i += 2) {
    vis[i] = false;
  }
  for (long long i = 3; i <= n; i += 2) {
    if (vis[i]) {
      prime.push_back(i);
      if (i > sqrt(n)) {
        continue;
      }
      for (long long j = i * i; j <= n; j += 2 * i) {
        vis[j] = false;
      }
    }
  }
}
int main() {
  sieve(100100);
  int t;
  cin >> t;
  for (int tc = 1; tc <= t; tc++) {
    long long p, q;
    cin >> p >> q;
    if (p % q != 0) {
      cout << p << '\n';
      continue;
    }
    vector<long long> factor, power;
    for (int i = 0; prime[i] <= sqrt(q); i++) {
      if (q % prime[i] == 0) {
        factor.push_back(prime[i]);
        long long cur = 0;
        while (q % prime[i] == 0) {
          cur++;
          q /= prime[i];
        }
        power.push_back(cur);
      }
    }
    if (q != 1) {
      factor.push_back(q);
      power.push_back(1);
    }
    long long ans = 0;
    for (int i = 0; i < factor.size(); i++) {
      long long r = factor[i];
      long long cur = p;
      while (cur % r == 0) {
        cur /= r;
      }
      for (long long j = 1; j < power[i]; j++) {
        cur *= r;
      }
      ans = max(ans, cur);
    }
    cout << ans << '\n';
  }
}

```