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 int maxn = 1e6 + 5;
const int inf = 0x3f3f3f3f;
const long long inF = 0x3f3f3f3f3f3f3f3f;
const long long mod = 1e9 + 7;
const double eps = 1e-8;
long long prime[maxn];
bool vis[maxn];
void getprime() {
  for (int i = 2; i < maxn; i++) {
    if (!vis[i]) prime[++prime[0]] = i;
    for (int j = 1; j <= prime[0] && i * prime[j] < maxn; j++) {
      vis[i * prime[j]] = 1;
      if (i % prime[j] == 0) break;
    }
  }
  return;
}
int main() {
  getprime();
  long long t, p, q;
  cin >> t;
  while (t--) {
    cin >> p >> q;
    if (p % q != 0) {
      cout << p << '\n';
      continue;
    }
    long long a;
    long long ans = 1;
    for (int i = 1; i <= prime[0]; i++) {
      if (q % prime[i] == 0) {
        int cnt = 0;
        a = p;
        while (q % prime[i] == 0) {
          cnt++;
          q /= prime[i];
        }
        while (a % prime[i] == 0) {
          a /= prime[i];
        }
        for (int j = 1; j < cnt; j++) a *= prime[i];
        ans = max(a, ans);
      }
    }
    if (q > 1) {
      while (p % q == 0) p /= q;
      ans = max(p, ans);
    }
    cout << ans << '\n';
  }
  return 0;
}
