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 N = 1e5;
vector<long long> prime;
int vis[N + 10];
long long qpow(long long a, long long b) {
  long long ans = 1, base = a;
  while (b) {
    if (b & 1) {
      ans *= base;
    }
    base *= base;
    b >>= 1;
  }
  return ans;
}
void getprime() {
  for (int i = 2; i <= N; ++i) {
    if (!vis[i]) {
      prime.push_back(i);
    }
    for (long long p : prime) {
      long long v = p * i;
      if (v > N) break;
      vis[v] = 1;
      if (i % p == 0) break;
    }
  }
}
int main() {
  ios::sync_with_stdio(0);
  int tt;
  cin >> tt;
  getprime();
  while (tt--) {
    vector<pair<long long, long long> > qyin;
    long long p, q;
    cin >> p >> q;
    if (p % q > 0) {
      cout << p << '\n';
      continue;
    }
    for (long long v : prime) {
      if (q % v == 0) {
        long long cnt = 0;
        while (q % v == 0) {
          q /= v;
          ++cnt;
        }
        qyin.push_back({v, cnt});
      }
    }
    if (q > 1) {
      qyin.push_back({q, 1});
    }
    long long ans = 0;
    for (auto& v : qyin) {
      long long tp = p;
      while (tp % v.first == 0) {
        tp /= v.first;
      }
      tp *= qpow(v.first, v.second - 1);
      ans = max(ans, tp);
    }
    cout << ans << '\n';
  }
}
