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>
const int maxn = 2e5 + 10;
const int mod = 998244353;
using namespace std;
template <typename T1, typename T2>
void ckmin(T1 &a, T2 b) {
  if (a > b) a = b;
}
template <typename T1, typename T2>
void ckmax(T1 &a, T2 b) {
  if (a < b) a = b;
}
template <typename T1, typename T2>
void addmo(T1 &a, T2 b) {
  a += b;
  if (a >= mod) a -= mod;
}
long long p, q;
void solve() {
  long long mi = 1e18;
  scanf("%lld%lld", &p, &q);
  long long pp = p;
  for (int i = 2; i * i <= q; i++) {
    long long tt = 1;
    while (q % i == 0) {
      tt *= i, q /= i;
    }
    if (p % tt) return printf("%lld\n", pp), void();
    long long tp = 1;
    while (p % i == 0) {
      tp *= i, p /= i;
    }
    if (tt > 1) ckmin(mi, tp / tt * i);
  }
  if (q > 1) {
    if (p % q) return printf("%lld\n", pp), void();
    long long tp = 1;
    while (p % q == 0) {
      tp *= q, p /= q;
    }
    ckmin(mi, tp);
  }
  printf("%lld\n", pp / mi);
}
signed main() {
  int T;
  scanf("%d", &T);
  while (T--) solve();
  return 0;
}
