#include <bits/stdc++.h>
using namespace std;
template <typename T, size_t N>
int SIZE(const T (&t)[N]) {
  return N;
}
template <typename T>
int SIZE(const T &t) {
  return t.size();
}
string to_string(const string s, int x1 = 0, int x2 = 1e9) {
  return '"' + ((x1 < s.size()) ? s.substr(x1, x2 - x1 + 1) : "") + '"';
}
string to_string(const char *s) { return to_string((string)s); }
string to_string(const bool b) { return (b ? "true" : "false"); }
string to_string(const char c) { return string({c}); }
template <size_t N>
string to_string(const bitset<N> &b, int x1 = 0, int x2 = 1e9) {
  string t = "";
  for (int __iii__ = min(x1, SIZE(b)), __jjj__ = min(x2, SIZE(b) - 1);
       __iii__ <= __jjj__; ++__iii__) {
    t += b[__iii__] + '0';
  }
  return '"' + t + '"';
}
template <typename A, typename... C>
string to_string(const A(&v), int x1 = 0, int x2 = 1e9, C... coords);
int l_v_l_v_l = 0, t_a_b_s = 0;
template <typename A, typename B>
string to_string(const pair<A, B> &p) {
  l_v_l_v_l++;
  string res = "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
  l_v_l_v_l--;
  return res;
}
template <typename A, typename... C>
string to_string(const A(&v), int x1, int x2, C... coords) {
  int rnk = rank<A>::value;
  string tab(t_a_b_s, ' ');
  string res = "";
  bool first = true;
  if (l_v_l_v_l == 0) res += '\n';
  res += tab + "[";
  x1 = min(x1, SIZE(v)), x2 = min(x2, SIZE(v));
  auto l = begin(v);
  advance(l, x1);
  auto r = l;
  advance(r, (x2 - x1) + (x2 < SIZE(v)));
  for (auto e = l; e != r; e = next(e)) {
    if (!first) {
      res += ", ";
    }
    first = false;
    l_v_l_v_l++;
    if (e != l) {
      if (rnk > 1) {
        res += '\n';
        t_a_b_s = l_v_l_v_l;
      };
    } else {
      t_a_b_s = 0;
    }
    res += to_string(*e, coords...);
    l_v_l_v_l--;
  }
  res += "]";
  if (l_v_l_v_l == 0) res += '\n';
  return res;
}
void dbgm() { ; }
template <typename Heads, typename... Tails>
void dbgm(Heads H, Tails... T) {
  cerr << to_string(H) << " | ";
  dbgm(T...);
}
bool vectorpair_sort_by_sec(const pair<int, int> &a, const pair<int, int> &b) {
  return a.second < b.second;
}
bool custom_compare(pair<long long, pair<int, int> > &a,
                    pair<long long, pair<int, int> > &b) {
  return a.first > b.first;
}
long long gcd(long long a, long long b) {
  if (b == 0) {
    return a;
  }
  return gcd(b, a % b);
}
long long _pow(long long a, long long b) {
  long long ans = 1;
  for (int i = 1; i <= b; i++) {
    ans = (ans * a);
  }
  return ans;
}
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
void testcase() {
  long long p, q;
  cin >> p >> q;
  if (p < q || p % q) {
    cout << p << "\n";
    return;
  }
  long long ans = 0;
  for (long long i = 1; i * i <= q; i++) {
    if (q % i == 0) {
      if (i != 1) {
        long long tmp = p;
        while (tmp % q == 0) tmp /= i;
        ans = max(ans, tmp);
      }
      long long tmp = p;
      while (tmp % q == 0) tmp /= q / i;
      ans = max(ans, tmp);
    }
  }
  cout << ans << "\n";
}
int main() {
  time_t t1, t2;
  t1 = clock();
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  ;
  int t;
  cin >> t;
  while (t--) testcase();
}

### Prompt

Please formulate a cpp solution to the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
template <typename T, size_t N>
int SIZE(const T (&t)[N]) {
  return N;
}
template <typename T>
int SIZE(const T &t) {
  return t.size();
}
string to_string(const string s, int x1 = 0, int x2 = 1e9) {
  return '"' + ((x1 < s.size()) ? s.substr(x1, x2 - x1 + 1) : "") + '"';
}
string to_string(const char *s) { return to_string((string)s); }
string to_string(const bool b) { return (b ? "true" : "false"); }
string to_string(const char c) { return string({c}); }
template <size_t N>
string to_string(const bitset<N> &b, int x1 = 0, int x2 = 1e9) {
  string t = "";
  for (int __iii__ = min(x1, SIZE(b)), __jjj__ = min(x2, SIZE(b) - 1);
       __iii__ <= __jjj__; ++__iii__) {
    t += b[__iii__] + '0';
  }
  return '"' + t + '"';
}
template <typename A, typename... C>
string to_string(const A(&v), int x1 = 0, int x2 = 1e9, C... coords);
int l_v_l_v_l = 0, t_a_b_s = 0;
template <typename A, typename B>
string to_string(const pair<A, B> &p) {
  l_v_l_v_l++;
  string res = "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
  l_v_l_v_l--;
  return res;
}
template <typename A, typename... C>
string to_string(const A(&v), int x1, int x2, C... coords) {
  int rnk = rank<A>::value;
  string tab(t_a_b_s, ' ');
  string res = "";
  bool first = true;
  if (l_v_l_v_l == 0) res += '\n';
  res += tab + "[";
  x1 = min(x1, SIZE(v)), x2 = min(x2, SIZE(v));
  auto l = begin(v);
  advance(l, x1);
  auto r = l;
  advance(r, (x2 - x1) + (x2 < SIZE(v)));
  for (auto e = l; e != r; e = next(e)) {
    if (!first) {
      res += ", ";
    }
    first = false;
    l_v_l_v_l++;
    if (e != l) {
      if (rnk > 1) {
        res += '\n';
        t_a_b_s = l_v_l_v_l;
      };
    } else {
      t_a_b_s = 0;
    }
    res += to_string(*e, coords...);
    l_v_l_v_l--;
  }
  res += "]";
  if (l_v_l_v_l == 0) res += '\n';
  return res;
}
void dbgm() { ; }
template <typename Heads, typename... Tails>
void dbgm(Heads H, Tails... T) {
  cerr << to_string(H) << " | ";
  dbgm(T...);
}
bool vectorpair_sort_by_sec(const pair<int, int> &a, const pair<int, int> &b) {
  return a.second < b.second;
}
bool custom_compare(pair<long long, pair<int, int> > &a,
                    pair<long long, pair<int, int> > &b) {
  return a.first > b.first;
}
long long gcd(long long a, long long b) {
  if (b == 0) {
    return a;
  }
  return gcd(b, a % b);
}
long long _pow(long long a, long long b) {
  long long ans = 1;
  for (int i = 1; i <= b; i++) {
    ans = (ans * a);
  }
  return ans;
}
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
void testcase() {
  long long p, q;
  cin >> p >> q;
  if (p < q || p % q) {
    cout << p << "\n";
    return;
  }
  long long ans = 0;
  for (long long i = 1; i * i <= q; i++) {
    if (q % i == 0) {
      if (i != 1) {
        long long tmp = p;
        while (tmp % q == 0) tmp /= i;
        ans = max(ans, tmp);
      }
      long long tmp = p;
      while (tmp % q == 0) tmp /= q / i;
      ans = max(ans, tmp);
    }
  }
  cout << ans << "\n";
}
int main() {
  time_t t1, t2;
  t1 = clock();
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  ;
  int t;
  cin >> t;
  while (t--) testcase();
}

```