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;
inline long long read() {
  char ch = getchar();
  long long p = 1, data = 0;
  while (ch < '0' || ch > '9') {
    if (ch == '-') p = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    data = data * 10 + (ch ^ 48);
    ch = getchar();
  }
  return p * data;
}
inline long long qpow(long long a, long long b) {
  long long r = 1;
  while (b) {
    if (b & 1) r = a * r;
    a = a * a;
    b >>= 1;
  }
  return r;
}
inline long long gcd(long long a, long long b) {
  if (!a || !b) return (!a) ? b : a;
  while (b ^= a ^= b ^= a %= b)
    ;
  return a;
}
inline long long lcm(long long a, long long b) { return a * b / gcd(a, b); }
void exgcd(long long a, long long b, long long &x, long long &y) {
  if (!b)
    x = 1, y = 0;
  else {
    exgcd(b, a % b, y, x);
    y -= x * (a / b);
  }
}
const int mod = 1e9 + 7;
long long a, b;
long long cal(long long x) {
  if (x == 1) return 1;
  long long ret = a;
  while (ret % b == 0) ret /= x;
  return ret;
}
int main() {
  int T = read();
  while (T--) {
    a = read(), b = read();
    if (a % b)
      printf("%lld\n", a);
    else {
      long long ans = cal(b);
      for (long long i = 2; i <= sqrt(b); i++) {
        if (b % i) continue;
        ans = max(ans, cal(i));
        ans = max(ans, cal(b / i));
      }
      printf("%lld\n", ans);
    }
  }
  return 0;
}
