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 int maxn = 5e7 + 10;
bool notprime[maxn];
int prime[maxn];
int cnt;
void getprime() {
  notprime[0] = notprime[1] = 1;
  cnt = 0;
  for (int i = 2; i < maxn; ++i) {
    if (!notprime[i]) {
      prime[cnt++] = i;
    }
    for (int j = 0; j < cnt && 1ll * i * prime[j] < maxn; ++j) {
      notprime[i * prime[j]] = 1;
    }
  }
}
long long p, q;
long long solve(long long a, int b) {
  long long res = 1;
  for (int i = 1; i <= b; i++) {
    res *= a;
  }
  return res;
}
long long quick(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = res * a;
    a = a * a;
    b >>= 1;
  }
  return res;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    cin >> p >> q;
    if (p % q != 0) {
      cout << p << endl;
      continue;
    }
    long long tmp = p;
    long long ans = 1e18;
    for (long long i = 2; i * i <= q; i++) {
      if (q % i == 0) {
        int num1 = 0, num2 = 0;
        while (q % i == 0) {
          q /= i;
          num1++;
        }
        while (p % i == 0) {
          p /= i;
          num2++;
        }
        ans = min(ans, quick(i, num2 - num1 + 1));
      }
    }
    if (q != 1) {
      int num = 0;
      while (p % q == 0) {
        p /= q;
        num++;
      }
      ans = min(ans, quick(q, num));
    }
    cout << tmp / ans << endl;
  }
}
