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 p1 = 1000000, p2 = 0;
char buf[1000005], wb[1000005];
long long getint() {
  long long ret = 0, flag = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') flag = -1;
    c = getchar();
  }
  while (c <= '9' && c >= '0') {
    ret = (ret << 3) + (ret << 1) + c - '0';
    c = getchar();
  }
  return ret * flag;
}
void pc(char x) {
  if (p2 >= 1000000) fwrite(wb, 1, 1000000, stdout), p2 = 0;
  wb[p2++] = x;
}
void wrt(long long x) {
  if (x < 0) pc('-'), x = -x;
  long long c[24] = {0};
  if (!x) return pc('0'), void();
  while (x) c[++c[0]] = x % 10, x /= 10;
  while (c[0]) pc(c[c[0]--] + '0');
}
long long fac[1005], num[1005], ans;
int main() {
  int Ti = getint();
  while (Ti--) {
    ans = 0;
    long long p = getint(), q = getint(), x = 0;
    fac[0] = 0;
    if (p % q > 0)
      wrt(p), pc('\n');
    else {
      long long sq = sqrt(q);
      for (int i = 2; i <= sq; i++) {
        if (q % i == 0) {
          fac[++fac[0]] = i;
          num[fac[0]] = 0;
          while (q % i == 0) ++num[fac[0]], q /= i;
          sq = sqrt(q);
        }
      }
      if (q > 1) fac[++fac[0]] = q, num[fac[0]] = 1;
      for (int i = 1; i <= fac[0]; i++) {
        long long tp = p;
        while (tp % fac[i] == 0) tp /= fac[i];
        for (int j = 1; j < num[i]; j++) tp = tp * fac[i];
        ans = max(ans, tp);
      }
      wrt(ans), pc('\n');
    }
  }
  fwrite(wb, 1, p2, stdout);
  return 0;
}
