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;
template <class T>
using V = vector<T>;
template <class T, class U>
using P = pair<T, U>;
using vll = V<ll>;
using vvll = V<vll>;
template <class T>
inline bool chmax(T& a, T b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
inline bool chmin(T& a, T b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
constexpr ll MOD = 1000000007;
constexpr ll HIGHINF = (ll)1e18;
constexpr int INF = 1e9;
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int t;
  cin >> t;
  while (t-- > 0) {
    ll p, q;
    cin >> p >> q;
    if (p % q != 0)
      cout << p << '\n';
    else {
      ll coef = p / q, tq = q;
      ll origc = coef;
      set<int> divq;
      for (ll i = 2; i * i <= tq; i++) {
        if (tq % i == 0) {
          divq.insert(i);
          while (tq % i == 0) tq /= i;
        }
      }
      if (tq > 1) divq.insert(tq);
      ll ans = 0;
      for (const int& dq : divq) {
        ll tmpq = q / dq;
        if (origc % dq != 0)
          chmax(ans, tmpq * origc);
        else {
          ll tc = origc;
          while (tc % dq == 0) tc /= dq;
          chmax(ans, tmpq * tc);
        }
      }
      cout << ans << '\n';
    }
  }
  return 0;
}
