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;
using ll = long long;
const int N = 1e1 + 4;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int t = 1;
  cin >> t;
  int cas = 0;
  while (t--) {
    ll n, m, i, j, cnt = 0, cnt1 = 0, cnt2 = 0, even = 0, odd = 0, len, k, r, l,
                   z = 0, x = 0, y = 0, flag = 0, sum = 0;
    ll a = 0, b = 0, c = 0, d = 0, ans = 0, rem, quot, zero = 0, fst = 0,
       null = 0, snd = 0, lst = 0, rone = 0, one = 0, pos = 0, neg = 0,
       mn = INT_MAX, mx = INT_MIN;
    char ch;
    int h1, h2, m1, m2, h;
    int velo1, velo2, ac1, ac2, tim, hour, mint, sec;
    int node, edge, u, v, cost;
    int bst, wrst;
    double nd, ad, bd, xd, sumd = 0.00;
    string str, str1 = "", str2 = "", str3 = "", strstore1 = "", strstore2 = "";
    cin >> n >> m;
    vector<ll> vec;
    k = sqrt(m);
    x = m;
    if (n % m || n < m) {
      cout << n << endl;
      continue;
    }
    for (i = 2; i <= sqrt(x); i++) {
      if (x % i == 0) {
        vec.push_back(i);
      }
      while (x % i == 0) {
        x /= i;
      }
    }
    if (x > 1) {
      vec.push_back(x);
    }
    sort(vec.begin(), vec.end());
    x = 1;
    for (i = 0; i < vec.size(); i++) {
      y = n;
      while (y % m == 0) {
        y /= vec[i];
      }
      x = max(x, y);
    }
    cout << x << endl;
  }
  return 0;
}
