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 MAX_N = 100 + 5;
const int MAX_M = 100000 + 5;
const int INF = 0x7fffffff;
const int inf = 1000000000;
const double EPS = 1e-6;
const unsigned long long base = 123;
const long long mod = 1e4 + 7;
const double pi = 4 * atan(1.0);
vector<int> v;
int vis[1000000];
int main() {
  int t;
  cin >> t;
  for (int i = 2; i * i <= 1000000000; i++) {
    if (!vis[i]) {
      v.push_back(i);
      for (int j = i * 2; j <= 100000; j += i) {
        vis[j] = 1;
      }
    }
  }
  while (t--) {
    long long p, q;
    map<long long, int> m1, m2;
    scanf("%lld%lld", &p, &q);
    if (p % q != 0) {
      cout << p << endl;
      continue;
    } else {
      for (int i = 0; i < v.size();) {
        if (q % v[i] == 0) {
          q /= v[i];
          m2[v[i]]++;
        } else {
          i++;
        }
      }
      if (q != 1) {
        m2[q]++;
      }
      long long mx = 0;
      for (auto i : m2) {
        long long u = 1;
        long long x = p;
        for (int j = 1; j <= i.second; j++) {
          u *= i.first;
        }
        while (x % u == 0) {
          x /= i.first;
        }
        mx = max(mx, x);
      }
      cout << mx << endl;
    }
  }
}
