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;
const int MAXN = 1e5 + 7;
long long zs[MAXN], zx[MAXN], tot = 0;
void sssai(int n) {
  for (int i = 2; i <= n; i++) {
    if (!zx[i]) {
      zs[++tot] = i;
      zx[i] = i;
    }
    for (int j = 1; j <= tot; j++) {
      if (i * zs[j] > n) break;
      zx[i * zs[j]] = zs[j];
      if (zx[i] == zs[j]) break;
    }
  }
}
int main() {
  sssai((int)1e5);
  int T;
  long long p, q;
  cin >> T;
  while (T--) {
    cin >> p >> q;
    long long res = p;
    if (p % q) {
      cout << p << endl;
      continue;
    }
    for (int i = 1; i <= tot; i++) {
      if (zs[i] > q) break;
      int cnt1 = 0, cnt2 = 0;
      while (q % zs[i] == 0) {
        cnt1++;
        q /= zs[i];
      }
      if (!cnt1) continue;
      long long t = p;
      while (t % zs[i] == 0) {
        cnt2++;
        t /= zs[i];
      }
      long long res2 = 1;
      for (int j = 1; j <= cnt2 - cnt1 + 1; j++) {
        res2 *= zs[i];
      }
      res = min(res, res2);
    }
    if (q != 1) {
      if (p % q == 0) {
        int cnt = 0;
        long long t = p;
        while (t % q == 0) {
          cnt++;
          t /= q;
        }
        long long res2 = 1;
        for (int i = 1; i <= cnt; i++) res2 *= q;
        res = min(res, res2);
      }
    }
    cout << p / res << endl;
  }
  return 0;
}
