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 long long mod = (long long)1e9 + 7;
const long long maxn = (long long)2e5 + 5;
const int nnn = 1048590;
const int inf = numeric_limits<int>::max() - 1;
const long long INF = 1e18;
long long dx[] = {0, 1, 0, -1};
long long dy[] = {1, 0, -1, 0};
long long dxx[] = {0, 1, 0, -1, 1, 1, -1, -1};
long long dyy[] = {1, 0, -1, 0, 1, -1, 1, -1};
void solve() {
  long long x, y;
  cin >> x >> y;
  if (x % y != 0) {
    cout << x << '\n';
    return;
  }
  vector<long long> v;
  long long p = x;
  vector<long long> r1, r2;
  for (long long i = 2; i * i <= y; i++) {
    if (y % i == 0) {
      long long cnt = 1;
      while (y % i == 0) y /= i, cnt *= i;
      v.push_back(i);
      r1.push_back(cnt);
      cnt = 1;
      while (p % i == 0) p /= i, cnt *= i;
      r2.push_back(cnt);
    }
  }
  if (y != 1) {
    v.push_back(y);
    r1.push_back(y);
    long long cnt = 1;
    while (p % y == 0) p /= y, cnt *= y;
    r2.push_back(cnt);
  }
  long long sz = v.size();
  long long ans = 1;
  for (long long i = 0; i < sz; i++) {
    ans = max(ans, (x / r2[i]) * (r1[i] / v[i]));
  }
  cout << ans << '\n';
  return;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  long long T;
  T = 1;
  cin >> T;
  long long CT = 0;
  while (T--) {
    solve();
  }
  return 0;
}
