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>
struct MI {
 private:
  char bb[1 << 14];
  FILE *f;
  char *bs, *be;
  char e;
  bool o, l;

 public:
  MI() : f(stdin), bs(0), be(0) {}
  inline char get() {
    if (o) {
      o = 0;
      return e;
    }
    if (bs == be) be = (bs = bb) + fread(bb, 1, sizeof(bb), f);
    if (bs == be) {
      l = 1;
      return -1;
    };
    return *bs++;
  }
  inline void unget(char c) {
    o = 1;
    e = c;
  }
  template <class T>
  inline T read() {
    T r;
    *this > r;
    return r;
  }
  template <class T>
  inline MI &operator>(T &);
};
template <class T>
struct Q {
  const static bool U = T(-1) >= T(0);
  inline void operator()(MI &t, T &r) const {
    r = 0;
    char c;
    bool y = 0;
    if (U)
      for (;;) {
        c = t.get();
        if (c == -1) goto E;
        if (isdigit(c)) break;
      }
    else
      for (;;) {
        c = t.get();
        if (c == -1) goto E;
        if (c == '-') {
          c = t.get();
          if (isdigit(c)) {
            y = 1;
            break;
          };
        } else if (isdigit(c))
          break;
        ;
      };
    for (;;) {
      if (c == -1) goto E;
      if (isdigit(c))
        r = r * 10 + (c ^ 48);
      else
        break;
      c = t.get();
    }
    t.unget(c);
  E:;
    if (y) r = -r;
  }
};
template <>
struct Q<char> {};
template <class T>
inline MI &MI::operator>(T &t) {
  Q<T>()(*this, t);
  return *this;
}
template <class T>
std::ostream &operator<(std::ostream &out, const T &t) {
  return out << t;
}
using std::cout;
MI cin;
template <typename T>
inline bool gmax(T &a, const T &b) {
  return a < b && (a = b, true);
}
long long p;
int q;
inline long long remove(int x) {
  long long t = p;
  while (!(t % x)) t /= x;
  return t;
}
inline long long work() {
  cin > p > q;
  if (p % q) return p;
  long long ans = 0;
  for (int i = 2; i * i <= q; ++i) {
    if (q % i) continue;
    long long t = 1;
    do {
      q /= i;
      t *= i;
    } while (!(q % i));
    gmax(ans, remove(i) * t / i);
  }
  if (q != 1) gmax(ans, remove(q));
  return ans;
}
int main() {
  for (int T = (cin.read<int>()); T; --T) cout < work() < ('\n');
}
