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;
long long t = 1, n, a, b, c[1000009], d[1000009];
map<long long, long long> mp2;
long long qp(long long x, long long y) {
  if (!y) return 1;
  long long p = qp(x, y / 2);
  if (y & 1) return p * p * x;
  return p * p;
}
int main() {
  cin >> t;
  while (t--) {
    long long c1 = 0, c2 = 0, ans, aa = 1e18, f = 0;
    cin >> a >> b, ans = a;
    for (long long i = 2; i <= sqrt(b); i += 1)
      if (!(b % i)) d[c2++] = i, mp2[i] += 1, b /= i, i -= 1;
    if (b > 1) d[c2++] = b, mp2[b] += 1;
    for (long long i = 0; i < c2; i += 1) {
      if (i && d[i] == d[i - 1]) continue;
      if (a % d[i]) {
        f = 1;
        break;
      }
      long long aaa = a, cur = 0;
      while (!(aaa % d[i])) aaa /= d[i], cur += 1;
      if (cur >= mp2[d[i]]) aa = min(aa, qp(d[i], cur - mp2[d[i]] + 1));
    }
    for (long long i = 0; i < c2; i += 1)
      if (a % d[i]) f = 1;
    for (long long i = 0; i < c2; i += 1) mp2[d[i]] = 0;
    if (f == 1)
      cout << ans << endl;
    else
      cout << (aa < 1e18 ? ans / aa : ans) << endl;
  }
  return 0;
}
