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 int64_t Nmax = 1e9 + 7;
const int64_t mod = 1000000007;
vector<int64_t> prime;
int64_t is_power_of_two(int64_t a) { return a && (!(a & (a - 1))); }
int64_t isperfectsquare(int64_t a) {
  double b = sqrt(a);
  return (b - ceil(b) == 0);
}
int64_t pow(int64_t a, int64_t b, int64_t m) {
  a = a % m;
  int64_t res = 1;
  while (b > 0) {
    if (b & 1) {
      res = res * a % m;
    }
    a = a * a % m;
    b = b / 2;
  }
  return res;
}
void seive(int64_t n) {
  bool isprime[n];
  memset(isprime, true, sizeof(isprime));
  prime.push_back(2);
  for (int64_t i = 3; i <= sqrt(n); i += 2) {
    if (isprime[i]) {
      for (int64_t j = i * i; j <= n; j += i) {
        isprime[j] = false;
      }
    }
  }
  for (int64_t i = 3; i <= n; i += 2) {
    if (isprime[i]) {
      prime.push_back(i);
    }
  }
}
vector<pair<int64_t, int64_t> > factors(int64_t n) {
  vector<pair<int64_t, int64_t> > fact_expo;
  for (int64_t i = 2; i <= sqrt(n); i++) {
    if (n % i == 0) {
      int64_t exp = 0;
      while (n % i == 0) {
        n = n / i;
        exp++;
      }
      fact_expo.push_back({i, exp});
    }
  }
  if (n > 1) {
    fact_expo.push_back({n, 1});
  }
  return fact_expo;
}
void solve() {
  int64_t n, k;
  cin >> n >> k;
  if (n % k) {
    cout << n << "\n";
    return;
  }
  int64_t ans = -1;
  vector<pair<int64_t, int64_t> > fact = factors(k);
  for (auto e : fact) {
    int64_t temp = n;
    while (temp % e.first == 0) {
      temp /= e.first;
    }
    ans = max(ans, (int64_t)(temp * pow(e.first, e.second - 1)));
  }
  cout << ans << "\n";
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int64_t t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
