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;
string to_string(char c) { return string(1, c); }
string to_string(bool b) { return b ? "true" : "false"; }
string to_string(const char* s) { return (string)s; }
string to_string(string s) { return s; }
template <class A>
string to_string(complex<A> c) {
  stringstream ss;
  ss << c;
  return ss.str();
}
string to_string(vector<bool> v) {
  string res = "{";
  for (int i = 0; i < (int)(v).size(); ++i) res += char('0' + v[i]);
  res += "}";
  return res;
}
template <size_t SZ>
string to_string(bitset<SZ> b) {
  string res = "";
  for (int i = 0; i < SZ; ++i) res += char('0' + b[i]);
  return res;
}
template <class A, class B>
string to_string(pair<A, B> p);
template <class T>
string to_string(T v) {
  bool fst = 1;
  string res = "{";
  for (const auto& x : v) {
    if (!fst) res += ", ";
    fst = 0;
    res += to_string(x);
  }
  res += "}";
  return res;
}
template <class A, class B>
string to_string(pair<A, B> p) {
  return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
}
void DBG() { cerr << "]" << '\n'; }
template <class H, class... T>
void DBG(H h, T... t) {
  cerr << to_string(h);
  if (sizeof...(t)) cerr << ", ";
  DBG(t...);
}
void EDBG(string names) { names = names; }
template <class H, class... T>
void EDBG(string names, H h, T... t) {
  auto pos = names.find(',');
  auto first_name = names.substr(0, pos);
  auto rest = names.substr(pos + 1);
  while (rest.front() == ' ') {
    rest = rest.substr(1);
  }
  cerr << "[" << first_name << "]: [" << to_string(h) << "]"
       << "\n";
  EDBG(rest, t...);
}
unsigned long long f(unsigned long long x, unsigned long long y) {
  unsigned long long fmax = 1;
  for (unsigned long long i = 2; i * i <= y; i++) {
    unsigned long long c = 0;
    while (y % i == 0) {
      c++;
      y /= i;
    }
    unsigned long long now = x;
    while (now % i == 0) {
      now /= i;
    }
    if (c > 1) fmax = max(fmax, now * (unsigned long long)pow(i, c - 1));
    if (c == 1) fmax = max(fmax, now);
    if (i > 2) i++;
  }
  if (y > 1) {
    unsigned long long now = x;
    while (now % y == 0) {
      now /= y;
    }
    fmax = max(fmax, now);
  }
  return fmax;
}
void solve() {
  unsigned long long n, m;
  cin >> n >> m;
  if (n % m != 0 or m > n)
    cout << n << '\n';
  else
    cout << f(n, m) << '\n';
}
int main() {
  int t = 1;
  cin >> t;
  while (t--) solve();
  return 0;
}
