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 sqrt_q_max = 1000000 + 5;
const int inf = INT_MAX;
const int mod = 998244353;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
map<int, int> mp;
long long _pow(int a, int b) {
  long long ret = 1;
  while (b > 0) {
    if (b & 1) ret = ret * a;
    a = a * a;
    b >>= 1;
  }
  return ret;
}
bool prime[sqrt_q_max];
map<int, int> pm;
long long solve(long long p, long long q) {
  mp.clear();
  if (p % q != 0) return p;
  long long p_tmp = p, q_tmp = q;
  for (auto it : pm) {
    int i = it.first;
    while (q_tmp % i == 0) {
      mp[i]++;
      q_tmp /= i;
    }
  }
  if (q_tmp != 1) mp[q_tmp]++;
  int t3 = clock();
  long long ans = 1;
  for (auto it : mp) {
    p_tmp = p;
    int prime = it.first, power = it.second;
    while (p_tmp % prime == 0) {
      p_tmp /= prime;
    }
    p_tmp *= _pow(prime, power - 1);
    ans = max(ans, p_tmp);
  }
  return ans;
}
void getPrime() {
  fill(prime, prime + sqrt_q_max, true);
  prime[1] = false;
  for (int i = 2; i * i <= sqrt_q_max; i++) {
    if (prime[i]) {
      for (int k = 2; i * k < sqrt_q_max; k++) {
        prime[i * k] = false;
      }
    }
  }
  for (int i = 2; i < sqrt_q_max; i++) {
    if (prime[i]) pm[i] = 1;
  }
}
int main() {
  int t;
  long long p, q;
  cin >> t;
  getPrime();
  for (int i = 1; i <= t; i++) {
    cin >> p >> q;
    cout << solve(p, q) << endl;
  }
}
