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 fact[1000001] = {0};
void preFact() {
  fact[0] = 1;
  fact[1] = 1;
  for (long long i = 2; i < 1000001; i++) {
    fact[i] = (fact[i - 1] * (i % 1000003)) % 1000003;
  }
}
bool isPrime(int n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (int i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
long long setBitNumber(long long n) {
  if (n == 0) return 0;
  long long msb = 0;
  n = n / 2;
  while (n != 0) {
    n = n / 2;
    msb++;
  }
  return (msb);
}
long long binaryExpon(long long a, long long b) {
  long long ans = 1;
  while (b > 0) {
    if (b % 2 == 1) {
      ans *= (a % 1000003) % 1000003;
      ans = ans % 1000003;
    }
    a *= (a % 1000003) % 1000003;
    a = a % 1000003;
    b = b >> 1;
  }
  return ans % 1000003;
}
long long nCr(long long n, long long r) {
  if (n < r) {
    return 0;
  } else {
    long long ans = fact[n];
    ans = ans * (binaryExpon(fact[r], (1000003 - 2)) % 1000003) % 1000003;
    ans = ans * (binaryExpon(fact[n - r], (1000003 - 2)) % 1000003) % 1000003;
    return ans;
  }
}
long long __gcd(long long a, long long b) {
  if (b == 0) {
    return a;
  }
  return __gcd(a % b, b);
}
void solve() {
  long long p;
  cin >> p;
  ;
  long long q;
  cin >> q;
  ;
  if (p % q != 0 || p < q) {
    cout << p << "\n";
    ;
    return;
  } else {
    vector<long long> factors;
    long long qq = q;
    for (long long i = 2; i * i <= q; i++) {
      if (qq % i == 0) {
        factors.push_back(i);
        while (qq % i == 0) {
          qq /= i;
        }
      }
    }
    if (qq > 1) {
      factors.push_back(qq);
    }
    long long finalans = -1;
    long long curr;
    for (auto x : factors) {
      long long pp = p;
      while (pp % q == 0) {
        pp /= x;
      }
      finalans = ((finalans > pp) ? finalans : pp);
    }
    cout << finalans << "\n";
    ;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  srand(chrono::high_resolution_clock::now().time_since_epoch().count());
  cin.tie(0);
  cout.tie(0);
  preFact();
  long long t;
  cin >> t;
  ;
  while (t--) {
    solve();
  }
}
