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;
long long mod = 1e9 + 7;
long long INF = 1e18;
bool isPrime(long long n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
long long nCr(long long n, long long r) {
  long long res = 1;
  if (n < r) return -1;
  r = min(r, n - r);
  for (long long i = 0; i < r; i++) {
    res *= (n - i);
    res /= (i + 1);
  }
  return res;
}
long long nPr(long long n, long long r) {
  long long res = 1;
  if (n < r) return -1;
  for (long long i = 0; i < r; i++) res *= (n - i);
  return res;
}
vector<long long> primes;
void sieve(long long n) {
  vector<bool> isp(n + 1, 1);
  isp[0] = 0;
  isp[1] = 0;
  for (long long i = 2; i * i <= n; i++) {
    if (isp[i] == 1) {
      for (long long j = i * i; j <= n; j += i) isp[j] = 0;
    }
  }
  for (long long i = 2; i <= n; i++) {
    if (isp[i]) primes.push_back(i);
  }
}
void solve() {
  long long p, q;
  cin >> p >> q;
  if (p % q != 0) {
    cout << p << "\n";
    return;
  }
  long long r = p;
  long long ans = 1;
  long long first = 0;
  vector<long long> fac;
  fac.push_back(p);
  for (long long i = 0; i < primes.size(); i++) {
    r = p;
    while (r % primes[i] == 0) {
      r /= primes[i];
      fac.push_back(r);
    }
  }
  sort(fac.begin(), fac.end());
  for (long long i = fac.size() - 1; i >= 0; i--) {
    if (fac[i] % q != 0) {
      cout << fac[i] << "\n";
      return;
    }
  }
  if (p % q != 0) {
    cout << p << "\n";
  } else if (q * q != p) {
    cout << p / q << "\n";
  } else
    cout << "1\n";
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t = 1;
  long long n = 2000000;
  sieve(n);
  cin >> t;
  while (t--) solve();
  return 0;
}
