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 long long N = 1000006, INF = 1e18, MOD = 1e9 + 7;
long long n, m, r, t, a[N], b[N], c[N], q = 1, e, o, u, k, x, y, ans, Ans[N],
                                        mx;
vector<long long> v[N];
string s, ss[N];
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0);
  cin >> q;
  while (q--) {
    ans = x = y = o = u = 0;
    e = INF;
    cin >> x >> y;
    if (x % y)
      cout << x << endl;
    else {
      k = sqrt(y);
      for (int i = 1; i <= k; i++) {
        if (y % i == 0) {
          if (i > 1) {
            u = x;
            while (u % y == 0) u /= i;
            ans = max(ans, u);
          }
          u = x;
          o = y / i;
          while (u % y == 0) u /= o;
          ans = max(ans, u);
        }
      }
      cout << max(ans, (long long)1) << endl;
    }
  }
}
