#include <bits/stdc++.h>
using namespace std;
inline int read() {
  char ch = getchar();
  int x = 0, cf = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') cf = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 3) + (x << 1) + (ch ^ 48);
    ch = getchar();
  }
  if (cf < 0)
    return -x;
  else
    return x;
}
int G[55];
template <class T>
inline void write(T x) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  int g = 0;
  do {
    G[++g] = x % 10;
    x /= 10;
  } while (x);
  for (int i = g; i >= 1; --i) putchar('0' + G[i]);
  putchar('\n');
}
long long f1[100010], f2[100010], sz1 = 0, sz2 = 0;
int main() {
  int t;
  cin >> t;
  while (t--) {
    sz1 = 0;
    sz2 = 0;
    long long q, p;
    cin >> p >> q;
    if (p % q) {
      cout << p << endl;
    } else {
      long long ovo = p / q;
      for (long long i = 1; i * i <= q; i++) {
        if (!(q % i)) {
          f1[++sz1] = i;
          if (q / i != i && q / i != q) {
            f1[++sz1] = q / i;
          }
        }
      }
      long long ans = INT_MIN;
      for (int i = 1; i <= sz1; i++) {
        long long t = ovo;
        while (!(t % (q / f1[i]))) {
          t /= (q / f1[i]);
        }
        ans = max(ans, t * f1[i]);
      }
      cout << ans << endl;
    }
  }
  return 0;
  cout << "UUZ ate it." << endl;
}

### Prompt

Your task is to create a Cpp solution 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;
inline int read() {
  char ch = getchar();
  int x = 0, cf = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') cf = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 3) + (x << 1) + (ch ^ 48);
    ch = getchar();
  }
  if (cf < 0)
    return -x;
  else
    return x;
}
int G[55];
template <class T>
inline void write(T x) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  int g = 0;
  do {
    G[++g] = x % 10;
    x /= 10;
  } while (x);
  for (int i = g; i >= 1; --i) putchar('0' + G[i]);
  putchar('\n');
}
long long f1[100010], f2[100010], sz1 = 0, sz2 = 0;
int main() {
  int t;
  cin >> t;
  while (t--) {
    sz1 = 0;
    sz2 = 0;
    long long q, p;
    cin >> p >> q;
    if (p % q) {
      cout << p << endl;
    } else {
      long long ovo = p / q;
      for (long long i = 1; i * i <= q; i++) {
        if (!(q % i)) {
          f1[++sz1] = i;
          if (q / i != i && q / i != q) {
            f1[++sz1] = q / i;
          }
        }
      }
      long long ans = INT_MIN;
      for (int i = 1; i <= sz1; i++) {
        long long t = ovo;
        while (!(t % (q / f1[i]))) {
          t /= (q / f1[i]);
        }
        ans = max(ans, t * f1[i]);
      }
      cout << ans << endl;
    }
  }
  return 0;
  cout << "UUZ ate it." << endl;
}

```