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;
int n, m, k, z, f[100039], s[100039], d[100039], head, dh, t, sh;
map<long long, int> q;
long long x, y, now, tot, pus, ans, flag;
int main() {
  register int i, j;
  for (i = 2; i * i <= 1e9; i++) {
    if (!f[i]) s[++head] = i;
    for (j = 1; j <= head && i * s[j] <= 100000; j++) {
      f[i * s[j]] = 1;
      if (i % s[j] == 0) break;
    }
  }
  scanf("%d", &t);
  while (t--) {
    scanf("%lld%lld", &x, &y);
    now = y;
    dh = sh = 0;
    pus = 1e18;
    for (i = 1; s[i] * s[i] <= y && i <= head; i++) {
      if (now % s[i] == 0) {
        d[++dh] = s[i];
        while (now % s[i] == 0) q[s[i]]++, now /= s[i], sh++;
      }
    }
    if (now != 1) q[now]++, d[++dh] = now, sh++;
    now = x;
    for (i = 1; i <= dh; i++) {
      if (x % d[i] == 0) {
        ans = 0;
        while (x % d[i] == 0) x /= d[i], ans++;
        if (ans >= q[d[i]]) {
          tot = 1;
          for (j = 1; j <= ans - q[d[i]] + 1; j++) tot *= d[i];
          pus = ((pus) < (tot) ? (pus) : (tot));
        }
        sh -= ((q[d[i]]) < (ans) ? (q[d[i]]) : (ans));
      }
    }
    if (sh)
      printf("%lld\n", now);
    else
      printf("%lld\n", now / pus);
    q.clear();
  }
}
