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 int read();
inline long long readl();
inline void write(int x, int ch);
inline void writel(long long x, int ch);
const int maxn = 1e7 + 10;
long long qpow(long long x, long long n) {
  long long res = 1;
  while (n) {
    if (n & 1) res = res * x;
    n >>= 1;
    x = x * x;
  }
  return res;
}
int a[110], b[110], n, x;
signed main() {
  int test = read();
  while (test--) {
    long long p = readl(), q = readl();
    if (p % q) {
      printf("%lld\n", p);
    } else {
      long long ans = 1;
      long long q1 = q;
      for (long long i = 2; i * i <= q; i++) {
        if (q1 % i == 0) {
          while (q1 % i == 0) {
            q1 /= i;
          }
          long long tmp = q / i;
          long long tt = p / q;
          while (tt % i == 0) {
            tt /= i;
          }
          ans = max(ans, tmp * tt);
        }
      }
      if (q1 > 1) {
        long long tmp = q / q1;
        long long tt = p / q;
        while (tt % q1 == 0) {
          tt /= q1;
        }
        ans = max(ans, tmp * tt);
      }
      writel(ans, 1);
    }
  }
}
inline void writel(long long x, int ch) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (x > 9) {
    writel(x / 10, 0);
  }
  putchar(x % 10 + '0');
  if (ch) printf("\n");
  return;
}
inline void write(int x, int ch) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (x > 9) {
    write(x / 10, 0);
  }
  putchar(x % 10 + '0');
  if (ch) printf("\n");
  return;
}
inline long long readl() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch > '9' || ch < '0') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch > '9' || ch < '0') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
