#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e6 + 15;
int prime[maxn];
bool visit[maxn] = {false};
vector<int> prim;
vector<int> nu;
void init() {
  int pos = 0;
  for (int i = 2; i < maxn; i++) {
    if (visit[i] == false) {
      prime[pos++] = i;
    }
    for (int j = 0; j < pos && prime[j] * i < maxn; j++) {
      visit[prime[j] * i] = true;
      if (i % prime[j] == 0) {
        break;
      }
    }
  }
}
bool isprime(long long test) {
  for (int i = 2; i <= sqrt(test); i++) {
    if (test / 2 == 0) {
      return false;
    }
  }
  return true;
}
void find(long long test) {
  for (int i = 0; prime[i] <= sqrt(test); i++) {
    int num = 0;
    while (test % prime[i] == 0) {
      num++;
      test /= prime[i];
    }
    if (num != 0) {
      prim.push_back(prime[i]);
      nu.push_back(num);
    }
  }
  if (isprime(test) && test > 1) {
    prim.push_back(test);
    nu.push_back(1);
  }
}
int main() {
  int t;
  init();
  cin >> t;
  long long p, q;
  while (t--) {
    cin >> p >> q;
    prim.clear();
    nu.clear();
    find(q);
    if (p % q != 0) {
      cout << p << endl;
    } else {
      long long maxans = -100;
      for (int i = 0; i < prim.size(); i++) {
        int tmpnum = 0;
        long long test = p;
        long long ans = p;
        while (test % prim[i] == 0) {
          test /= prim[i];
          tmpnum++;
        }
        int reducenum = tmpnum - nu[i] + 1;
        ans = ans / pow(prim[i], reducenum);
        maxans = max(ans, maxans);
      }
      cout << maxans << endl;
    }
  }
}

### 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;
const int maxn = 1e6 + 15;
int prime[maxn];
bool visit[maxn] = {false};
vector<int> prim;
vector<int> nu;
void init() {
  int pos = 0;
  for (int i = 2; i < maxn; i++) {
    if (visit[i] == false) {
      prime[pos++] = i;
    }
    for (int j = 0; j < pos && prime[j] * i < maxn; j++) {
      visit[prime[j] * i] = true;
      if (i % prime[j] == 0) {
        break;
      }
    }
  }
}
bool isprime(long long test) {
  for (int i = 2; i <= sqrt(test); i++) {
    if (test / 2 == 0) {
      return false;
    }
  }
  return true;
}
void find(long long test) {
  for (int i = 0; prime[i] <= sqrt(test); i++) {
    int num = 0;
    while (test % prime[i] == 0) {
      num++;
      test /= prime[i];
    }
    if (num != 0) {
      prim.push_back(prime[i]);
      nu.push_back(num);
    }
  }
  if (isprime(test) && test > 1) {
    prim.push_back(test);
    nu.push_back(1);
  }
}
int main() {
  int t;
  init();
  cin >> t;
  long long p, q;
  while (t--) {
    cin >> p >> q;
    prim.clear();
    nu.clear();
    find(q);
    if (p % q != 0) {
      cout << p << endl;
    } else {
      long long maxans = -100;
      for (int i = 0; i < prim.size(); i++) {
        int tmpnum = 0;
        long long test = p;
        long long ans = p;
        while (test % prim[i] == 0) {
          test /= prim[i];
          tmpnum++;
        }
        int reducenum = tmpnum - nu[i] + 1;
        ans = ans / pow(prim[i], reducenum);
        maxans = max(ans, maxans);
      }
      cout << maxans << endl;
    }
  }
}

```