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 int MOD = 1E18 + 7;
long long int gcd(long long int a, long long int b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
long long int gcd_ex(long long int a, long long int b, long long int& x,
                     long long int& y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  }
  long long int x1, y1;
  long long int g = gcd_ex(b, a % b, x1, y1);
  x = y1;
  y = x1 - y1 * (a / b);
  return g;
}
long long int multi_mod_inv(long long int a, long long int m) {
  long long int x, y;
  long long int g = gcd_ex(a, m, x, y);
  if (g != 1) {
    return -1;
  }
  x = (x % m + m) % m;
  return x;
}
long long int mod_div(long long int a, long long int b, long long int m) {
  if (multi_mod_inv(b, m) != -1) {
    return (a % m * multi_mod_inv(b, m) % m) % m;
  } else {
    return -1;
  }
}
long long int lcm(long long int a, long long int b) {
  return a / gcd(a, b) * b;
}
long long int power(long long int a, long long int n) {
  if (n == 0) return 1;
  long long int p = power(a, n / 2);
  p = (p * p) % MOD;
  if (n % 2)
    return (p * a) % MOD;
  else
    return p % MOD;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  long long int test = 1;
  cin >> test;
  while (test--) {
    long long int p, q;
    cin >> p >> q;
    if (p % q != 0) {
      cout << p << '\n';
    } else {
      vector<pair<long long int, long long int>> div;
      long long int temp = q;
      for (long long int i = 2; i * i <= temp; i++) {
        long long int count = 0;
        while (temp % i == 0) {
          temp /= i;
          count++;
        }
        if (count) {
          div.emplace_back(i, count);
        }
      }
      if (temp > 1) {
        div.emplace_back(temp, 1);
      }
      long long int ans = 0;
      for (auto x : div) {
        long long int temp_p = p;
        while (temp_p % q == 0) {
          temp_p /= x.first;
        }
        ans = max(ans, temp_p);
      }
      cout << ans << '\n';
    }
  }
  return 0;
}
