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;
const long long N = 1e3 + 5;
long long t, p, q, cnt, d[N], a[N], res, tmp;
long long mul(long long x, long long n) {
  if (n < 1) return 1;
  long long ans = 1;
  for (; n; n >>= 1, x = x * x)
    if (n & 1) ans = ans * x;
  return ans;
}
void solve(long long x) {
  for (long long i = 2; i <= sqrt(x); i++) {
    if (x % i != 0) continue;
    d[++cnt] = i;
    while (x % i == 0) x /= i, a[cnt]++;
  }
  if (x != 1) d[++cnt] = x, a[cnt]++;
}
long long work(long long x, long long y) {
  long long res = 0;
  while (x % y == 0) x /= y, res++;
  return res;
}
signed main() {
  scanf("%lld", &t);
  while (t--) {
    scanf("%lld%lld", &p, &q);
    if (q > p) {
      printf("%lld\n", p);
      continue;
    }
    cnt = 0, solve(q), res = p, tmp = 0;
    for (long long i = 1; i <= cnt; i++)
      if (p % d[i] == 0)
        res = min(res, mul(d[i], work(p, d[i]) - a[i] + 1)), tmp++;
    if (tmp != cnt) res = 1;
    printf("%lld\n", p / res);
    for (long long i = 1; i <= cnt; i++) d[i] = a[i] = 0;
  }
  return 0;
}
