#include <bits/stdc++.h>
using namespace std;
ifstream in("test.in");
ofstream out("test.out");
int T;
long long Pow(long long exp, long long n) {
  long long rasp = 1;
  while (exp > 0) {
    if (exp % 2)
      rasp *= n, n *= n;
    else
      n *= n;
    exp /= 2;
  }
  return rasp;
}
long long fp1[2][1000], fp2[2][1000];
int fp(long long a, long long fp[][1000]) {
  long long div = 2;
  int poz = 0;
  while (div * div <= a) {
    int cnt = 0;
    while (a % div == 0) {
      cnt++;
      a /= div;
    }
    if (cnt > 0) {
      fp[0][poz] = div;
      fp[1][poz] = cnt;
      poz++;
    }
    div++;
  }
  if (a > 1) {
    fp[0][poz] = a;
    fp[1][poz] = 1;
    poz++;
  }
  return poz;
}
int main() {
  cin >> T;
  long long a, b, u;
  for (int i = 1; i <= T; i++) {
    cin >> a >> b;
    u = a;
    int pozCom = -1;
    long long minim = 1e18 - 1, raspuns = a, c = a;
    if (a < b || a % b != 0) {
      cout << a << '\n';
      continue;
    }
    int len2, gasit = 0;
    len2 = fp(b, fp2), len2--;
    for (int j = 0; j <= len2; j++) {
      long long val = fp2[0][j], cnt = 0;
      if (a % val == 0) {
        while (c % val == 0) {
          cnt++;
          c /= val;
        }
        if (cnt < fp2[1][j]) gasit = 1;
        if (cnt >= fp2[1][j] && Pow(cnt - fp2[1][j] + 1, fp2[0][j]) <= minim &&
            gasit == 0) {
          minim = Pow(cnt - fp2[1][j] + 1, fp2[0][j]);
          pozCom = j;
        }
      }
    }
    if (pozCom != -1 && gasit != 1) a /= minim;
    if (u % a != 0 || (u % a == 0 and a % b == 0)) a = 1;
    cout << a << '\n';
  }
  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;
ifstream in("test.in");
ofstream out("test.out");
int T;
long long Pow(long long exp, long long n) {
  long long rasp = 1;
  while (exp > 0) {
    if (exp % 2)
      rasp *= n, n *= n;
    else
      n *= n;
    exp /= 2;
  }
  return rasp;
}
long long fp1[2][1000], fp2[2][1000];
int fp(long long a, long long fp[][1000]) {
  long long div = 2;
  int poz = 0;
  while (div * div <= a) {
    int cnt = 0;
    while (a % div == 0) {
      cnt++;
      a /= div;
    }
    if (cnt > 0) {
      fp[0][poz] = div;
      fp[1][poz] = cnt;
      poz++;
    }
    div++;
  }
  if (a > 1) {
    fp[0][poz] = a;
    fp[1][poz] = 1;
    poz++;
  }
  return poz;
}
int main() {
  cin >> T;
  long long a, b, u;
  for (int i = 1; i <= T; i++) {
    cin >> a >> b;
    u = a;
    int pozCom = -1;
    long long minim = 1e18 - 1, raspuns = a, c = a;
    if (a < b || a % b != 0) {
      cout << a << '\n';
      continue;
    }
    int len2, gasit = 0;
    len2 = fp(b, fp2), len2--;
    for (int j = 0; j <= len2; j++) {
      long long val = fp2[0][j], cnt = 0;
      if (a % val == 0) {
        while (c % val == 0) {
          cnt++;
          c /= val;
        }
        if (cnt < fp2[1][j]) gasit = 1;
        if (cnt >= fp2[1][j] && Pow(cnt - fp2[1][j] + 1, fp2[0][j]) <= minim &&
            gasit == 0) {
          minim = Pow(cnt - fp2[1][j] + 1, fp2[0][j]);
          pozCom = j;
        }
      }
    }
    if (pozCom != -1 && gasit != 1) a /= minim;
    if (u % a != 0 || (u % a == 0 and a % b == 0)) a = 1;
    cout << a << '\n';
  }
  return 0;
}

```