#include <bits/stdc++.h>
using namespace std;
const string here = "here\n";
const string yes = "YES\n";
const string no = "NO\n";
const long long M = 1e9 + 7;
const long double PI = 2.0 * acos(0);
long long binpow(long long a, long long b) {
  long long res = 1;
  for (; b; b >>= 1ll) {
    if (b & 1) res = (1ll * res * a);
    a = (1ll * a * a);
  }
  return res;
}
void pre() {}
void solve() {
  long long p, q, k, i, j, l, r;
  bool flag;
  cin >> p >> q;
  if (p % q) {
    cout << p << '\n';
    return;
  }
  long long x = q;
  set<long long> s;
  map<long long, long long> mp;
  for (long long i = 2; i * i <= x; ++i) {
    if (x % i == 0) {
      s.insert(i);
      while (x % i == 0) {
        x /= i;
        mp[i]++;
      }
    }
  }
  if (x > 1) {
    s.insert(x);
    mp[x]++;
  }
  long long ans = 1;
  for (long long pr : s) {
    long long y = binpow(pr, mp[pr] - 1);
    k = p;
    while (k % pr == 0) k /= pr;
    ans = max(ans, y * k);
  }
  cout << ans << '\n';
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  pre();
  long long t, _;
  cin >> t;
  _ = t;
  while (t--) solve();
  return 0;
}

### Prompt

Please create a solution in Cpp 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;
const string here = "here\n";
const string yes = "YES\n";
const string no = "NO\n";
const long long M = 1e9 + 7;
const long double PI = 2.0 * acos(0);
long long binpow(long long a, long long b) {
  long long res = 1;
  for (; b; b >>= 1ll) {
    if (b & 1) res = (1ll * res * a);
    a = (1ll * a * a);
  }
  return res;
}
void pre() {}
void solve() {
  long long p, q, k, i, j, l, r;
  bool flag;
  cin >> p >> q;
  if (p % q) {
    cout << p << '\n';
    return;
  }
  long long x = q;
  set<long long> s;
  map<long long, long long> mp;
  for (long long i = 2; i * i <= x; ++i) {
    if (x % i == 0) {
      s.insert(i);
      while (x % i == 0) {
        x /= i;
        mp[i]++;
      }
    }
  }
  if (x > 1) {
    s.insert(x);
    mp[x]++;
  }
  long long ans = 1;
  for (long long pr : s) {
    long long y = binpow(pr, mp[pr] - 1);
    k = p;
    while (k % pr == 0) k /= pr;
    ans = max(ans, y * k);
  }
  cout << ans << '\n';
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  pre();
  long long t, _;
  cin >> t;
  _ = t;
  while (t--) solve();
  return 0;
}

```