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;
mt19937 rnd(chrono::high_resolution_clock::now().time_since_epoch().count());
mt19937_64 rnd64(
    chrono::high_resolution_clock::now().time_since_epoch().count());
long long power(long long a, long long m, long long mod) {
  long long ans = 1;
  while (m) {
    if (m % 2) ans *= a, ans %= (mod);
    a = (a * a) % mod;
    m >>= 1;
  }
  return ans;
}
struct custom_hash {
  static uint64_t splitmix64(uint64_t x) {
    x += 0x9e3779b97f4a7c15;
    x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
    x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
    return x ^ (x >> 31);
  }
  size_t operator()(uint64_t x) const {
    static const uint64_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(x + FIXED_RANDOM);
  }
  size_t operator()(pair<int64_t, int64_t> Y) const {
    static const uint64_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(Y.first * 31 + Y.second + FIXED_RANDOM);
  }
};
const long long inf = 1e18;
const long long M = 1000000007;
void solve() {
  long long i, j, _;
  long long p, q;
  cin >> p >> q;
  if (p % q) {
    cout << p << '\n';
    return;
  }
  long long ans = p, tmp = p;
  for (int i = 2; i * i <= q; ++i) {
    long long c = 0;
    if (q % i) continue;
    while (q % i == 0) q /= i, --c;
    while (p % i == 0) p /= i, ++c;
    ++c;
    ans = min(ans, (long long)pow(i, c));
  }
  long long c = -1;
  while (q > 1 && p % q == 0) p /= q, ++c;
  ++c;
  if (q > 1) ans = min(ans, (long long)pow(q, c));
  cout << tmp / ans << '\n';
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long T = 1;
  cin >> T;
  while (T--) {
    solve();
  }
  return 0;
}
