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 i_inf = 1e9;
const long long inf = 2e18;
const long long mod = 1e9 + 7;
const long double eps = 1e-13;
const long double pi = 3.14159265359;
const int mxn = 5e5 + 5;
mt19937 _rand(time(NULL));
clock_t z;
long long modpow(long long x, long long n) {
  if (n == 0) return 1 % mod;
  long long u = modpow(x, n / 2);
  u = (u * u) % mod;
  if (n % 2) u = (u * x) % mod;
  return u;
}
long long inverse(long long x) { return modpow(x, mod - 2); }
long long _add(long long ans, long long val) {
  ans = (ans + val) % mod;
  return ans;
}
long long mul(long long ans, long long val) {
  ans = (ans * val) % mod;
  return ans;
}
long long sub(long long ans, long long val) {
  ans = (ans - val) % mod;
  if (ans < 0) ans += mod;
  return ans;
}
long long _div(long long ans, long long val) {
  ans = (ans * inverse(val)) % mod;
  return ans;
}
long long fakt[mxn];
long long bc(long long n, long long k) {
  if (n < k) return inverse(0);
  return (fakt[n] * inverse((fakt[k] * fakt[n - k]) % mod)) % mod;
}
void solve() {
  long long p, q;
  cin >> p >> q;
  if (p % q != 0) {
    cout << p << endl;
    return;
  }
  long long m = q;
  vector<pair<long long, long long> > v;
  for (long long i = 2; i * i <= m; i++) {
    int cnt = 0;
    while (m % i == 0) {
      ++cnt;
      m /= i;
    }
    if (cnt > 0) v.push_back({i, cnt});
  }
  if (m > 1) v.push_back({m, 1});
  long long ans = inf;
  for (auto u : v) {
    long long prime = u.first;
    long long n = p;
    int cnt = 0;
    while (n % prime == 0) {
      ++cnt;
      n /= prime;
    }
    long long tmp = 1;
    for (int i = (0); i < (cnt - u.second + 1); i++) tmp *= prime;
    ans = min(ans, tmp);
  }
  cout << p / ans << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
}
