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;
double PI = acos(-1.0);
double eps = 1e-9;
int_fast64_t L = 31250001;
const int_fast64_t md = 1e9 + 7;
vector<int_fast64_t> check;
vector<int_fast64_t> prime;
int main(void) {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  L = 31623;
  check.assign(495, 0);
  check[0 >> 6] |= (1LL << (0 & 63));
  check[1 >> 6] |= (1LL << (1 & 63));
  prime.push_back(2);
  for (int_fast64_t i = 4; i <= L; i += 2) check[i >> 6] |= (1LL << (i & 63));
  for (int_fast64_t i = 3; i * i <= L; i += 2)
    if (!(check[i >> 6] & (1LL << (i & 63))))
      for (int_fast64_t j = i * i; j <= L; j += 2 * i)
        check[j >> 6] |= (1LL << (j & 63));
  for (int_fast64_t i = 3; i <= L; i += 2)
    if (!(check[i >> 6] & (1LL << (i & 63)))) prime.push_back(i);
  int_fast64_t t;
  cin >> t;
  while (t--) {
    int_fast64_t p, q;
    cin >> p >> q;
    if (p % q) {
      cout << p << '\n';
      continue;
    }
    int_fast64_t ans = 1;
    for (auto &i : prime) {
      if (i * i > q) break;
      if (!(q % i)) {
        int_fast64_t temp = p, cnt = 0;
        while (!(q % i)) {
          q /= i;
          ++cnt;
        }
        while (!(temp % i)) temp /= i;
        for (int_fast64_t x = 1; x < cnt; ++x) temp *= i;
        ans = max(ans, temp);
      }
    }
    if (q > 1) {
      while (!(p % q)) p /= q;
      ans = max(ans, p);
    }
    cout << ans << '\n';
  }
  return 0;
}
