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.
#include <bits/stdc++.h>
using namespace std;
inline int read() {
  int x = 0, neg = 1;
  char op = getchar();
  while (!isdigit(op)) {
    if (op == '-') neg = -1;
    op = getchar();
  }
  while (isdigit(op)) {
    x = 10 * x + op - '0';
    op = getchar();
  }
  return neg * x;
}
inline void print(int x) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (x >= 10) print(x / 10);
  putchar(x % 10 + '0');
}
bool cmp(int x, int y) { return x > y; }
int gcd(int a, int b) {
  if (b == 0) return a;
  return (b, a % b);
}
vector<int> y;
int main() {
  ios::sync_with_stdio(false);
  int t;
  cin >> t;
  while (t--) {
    y.clear();
    long long q, p;
    cin >> p >> q;
    if ((p < q) || (p > q && p % q)) {
      cout << p << '\n';
    } else {
      long long temp = q;
      for (int i = 2; i * i <= q; ++i) {
        while (temp % i == 0) {
          y.push_back(i);
          temp /= i;
        }
      }
      if (temp > 1) y.push_back(temp);
      long long anstemp = p, ans = 0;
      for (int i = 0; i < y.size(); ++i) {
        anstemp = p;
        while (anstemp % q == 0 && anstemp % y[i] == 0) {
          anstemp /= y[i];
        }
        ans = max(anstemp, ans);
      }
      cout << ans << '\n';
    }
  }
  return 0;
}
