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 = 21 * 200001;
const long long INF = 1000000000ll;
const long long LLINF = 1000000000000000000ll;
const int MOD = 998244353;
int main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  cout.precision(10);
  cout << fixed;
  int t;
  cin >> t;
  for (int tt = 0; tt < (int)(t); tt++) {
    long long p, q;
    cin >> p >> q;
    if (p % q != 0) {
      cout << p << "\n";
      continue;
    }
    vector<long long> qDivs;
    vector<long long> qCnt;
    vector<long long> pCnt;
    long long qq = q;
    long long pp = p;
    long long k = 2;
    while (k * k <= qq) {
      if (qq % k == 0) {
        qDivs.push_back(k);
        qCnt.push_back(1);
        pCnt.push_back(1);
        while (qq % k == 0) {
          qq /= k;
          qCnt.back() *= k;
        }
        while (pp % k == 0) {
          pp /= k;
          pCnt.back() *= k;
        }
      }
      k++;
    }
    if (qq > 1) {
      qDivs.push_back(qq);
      qCnt.push_back(qq);
      pCnt.push_back(1);
      while (pp % qq == 0) {
        pp /= qq;
        pCnt.back() *= qq;
      }
    }
    long long res = 0;
    for (int i = 0; i < (int)(qDivs.size()); i++) {
      res = max(res, p / pCnt[i] * qCnt[i] / qDivs[i]);
    }
    cout << res << "\n";
  }
  return 0;
}
