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 long long mx = 200001;
const long long mod = 1e9 + 7;
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long extgcd(long long a, long long b, long long& x, long long& y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  }
  long long x1, y1;
  long long d = extgcd(b, a % b, x1, y1);
  x = y1;
  y = x1 - y1 * (a / b);
  return d;
}
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;
}
bool isPowerOfTwo(long long n) {
  if (n == 0) return false;
  return (ceil(log2(n)) == floor(log2(n)));
}
long long powm(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return res;
}
long long divide(long long a, long long b) {
  return (a % mod) * powm(b, mod - 2) % mod;
}
long long mul(long long a, long long b) {
  return ((a % mod) * (b % mod)) % mod;
}
long long add(long long a, long long b) { return (a % mod + b % mod) % mod; }
long long Totfun(long long n) {
  long long z = n;
  if (n % 2 == 0) {
    while (n % 2 == 0) {
      n /= 2;
    }
    z /= 2;
  }
  for (long long i = 3; i <= sqrt(n); i += 2) {
    if (isPrime(i) && n % i == 0) {
      while (n % i == 0) {
        n /= i;
      }
      z -= z / i;
    }
  }
  if (n > 1) {
    z -= z / n;
  }
  return z;
}
long long fact[100001];
long long nCr(long long n, long long r) {
  if (r > n || r < 0) return 0;
  long long z = fact[n];
  z = mul(z, powm(fact[n - r], mod - 2));
  z = mul(z, powm(fact[r], mod - 2));
  return z;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t;
  cin >> t;
  while (t--) {
    long long p, q;
    cin >> p >> q;
    if (p % q)
      cout << p << "\n";
    else {
      vector<pair<long long, long long> > z;
      for (long long i = 2; i < (long long)sqrt(q) + 1; i++) {
        if (q % i == 0 && isPrime(i)) {
          long long x = 1;
          while (q % i == 0) {
            x--;
            q /= i;
          }
          long long y = p;
          while (y % i == 0) {
            x++;
            y /= i;
          }
          z.push_back({i, x});
        }
      }
      if (q > 1) {
        long long x = 0;
        long long y = p;
        while (y % q == 0) {
          y /= q;
          x++;
        }
        z.push_back({q, x});
      }
      long long ans = 1;
      for (long long i = 0; i < ((long long)z.size()); i++) {
        long long x = 1;
        for (long long j = 0; j < z[i].second; j++) x *= z[i].first;
        ans = ((ans > p / x) ? ans : p / x);
      }
      cout << ans << "\n";
    }
  }
  return 0;
}
