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>
const long long N = 1e6 + 10;
using namespace std;
long long p, q;
template <class I>
inline void read(I &s) {
  char c = getchar();
  for (; !isdigit(c); c = getchar())
    ;
  for (s = 0; isdigit(c); c = getchar()) s = s * 10 + c - 48;
}
template <class I>
inline void write(I x) {
  (x < 0 ? x = ~x + 1, putchar('-') : 0), (x > 9 ? write(x / 10) : void());
  putchar(x % 10 + 48);
}
long long pr[N], cnt = 0, vis[N], c[N];
void Euler(long long n) {
  vis[1] = 1;
  for (long long i = 2; i <= n; ++i) {
    if (!vis[i]) pr[++cnt] = i;
    for (long long j = 1; j <= cnt && pr[j] * i <= N - 10; ++j)
      if (i % pr[j])
        vis[pr[j] * i] = 1;
      else
        break;
  }
}
inline long long Power(long long a, long long b) {
  long long ans = 1;
  for (; b; b >>= 1) (b & 1 ? ans = ans * a : 0), a = a * a;
  return ans;
}
void Work() {
  read(p), read(q);
  long long tot = 0, P[N];
  for (long long j = 1; pr[j] * pr[j] <= q && j <= cnt; ++j) {
    long long sum = 0;
    while (q % pr[j] == 0) ++sum, q /= pr[j];
    if (sum > 0) P[++tot] = pr[j], c[tot] = sum;
  }
  if (q > 1) P[++tot] = q, c[tot] = 1;
  long long ans = 0;
  for (long long i = 1; i <= tot; ++i) {
    long long tmp = p, t = 0;
    while (tmp % P[i] == 0) tmp /= P[i], ++t;
    if (t + 1 > c[i]) {
      long long nmsl = p;
      for (long long j = 1; j <= t - c[i] + 1; ++j) nmsl /= P[i];
      ans = max(ans, nmsl);
    } else
      ans = max(ans, p);
  }
  write(ans), putchar(10);
}
signed main(void) {
  long long T;
  read(T), Euler(N - 10);
  while (T--) Work();
  return 0;
}
