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;
constexpr int INF = 0x3f3f3f3f;
constexpr ll LINF = 0x3f3f3f3f3f3f3f3fLL;
constexpr double EPS = 1e-8;
constexpr int MOD = 1000000007;
constexpr int dy[] = {1, 0, -1, 0}, dx[] = {0, -1, 0, 1};
constexpr int dy8[] = {1, 1, 0, -1, -1, -1, 0, 1},
              dx8[] = {0, -1, -1, -1, 0, 1, 1, 1};
template <typename T, typename U>
inline bool chmax(T &a, U b) {
  return a < b ? (a = b, true) : false;
}
template <typename T, typename U>
inline bool chmin(T &a, U b) {
  return a > b ? (a = b, true) : false;
}
struct IOSetup {
  IOSetup() {
    cin.tie(nullptr);
    ios_base::sync_with_stdio(false);
    cout << fixed << setprecision(20);
  }
} iosetup;
template <typename T>
std::vector<std::pair<T, int>> prime_factorization(T val) {
  std::vector<std::pair<T, int>> res;
  for (T i = 2; i * i <= val; ++i) {
    if (val % i != 0) continue;
    int exponent = 0;
    while (val % i == 0) {
      ++exponent;
      val /= i;
    }
    res.emplace_back(i, exponent);
  }
  if (val != 1) res.emplace_back(val, 1);
  return res;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    ll p, q;
    cin >> p >> q;
    if (p % q > 0) {
      cout << p << '\n';
    } else {
      ll ans = 1;
      for (auto [k, v] : prime_factorization(q)) {
        ll tmp = p;
        while (tmp % k == 0) tmp /= k;
        for (int _ = (0); _ < (v - 1); ++_) tmp *= k;
        chmax(ans, tmp);
      }
      cout << ans << '\n';
    }
  }
  return 0;
}
