#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using ld = long double;
ll power(ll a, ll b) {
  ll ret = 1;
  while (b) {
    if (b & 1) {
      ret *= a;
    }
    b >>= 1;
    a *= a;
  }
  return ret;
}
vector<pll> gogo(ll x) {
  ll y = x;
  vector<pll> ret;
  for (ll i = 2; i * i <= x; i++) {
    if (y % i == 0) {
      ll cnt = 0;
      while (y % i == 0) y /= i, cnt++;
      ret.emplace_back(i, cnt);
    }
  }
  if (y != 1) ret.emplace_back(y, 1);
  return ret;
}
void solve() {
  ll p, q;
  cin >> p >> q;
  if (p % q) {
    cout << p << '\n';
    return;
  }
  auto get = gogo(q);
  ll ans = 0;
  for (int i = 0; i < get.size(); i++) {
    ll su1 = power(get[i].first, get[i].second);
    ll t = p;
    while (t % get[i].first == 0) t /= get[i].first;
    t *= su1 / get[i].first;
    ans = max(ans, t);
  }
  cout << ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  int t;
  cin >> t;
  while (t--) solve();
}

### Prompt

Develop a solution in cpp to the problem described below:

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;
using ll = long long;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using ld = long double;
ll power(ll a, ll b) {
  ll ret = 1;
  while (b) {
    if (b & 1) {
      ret *= a;
    }
    b >>= 1;
    a *= a;
  }
  return ret;
}
vector<pll> gogo(ll x) {
  ll y = x;
  vector<pll> ret;
  for (ll i = 2; i * i <= x; i++) {
    if (y % i == 0) {
      ll cnt = 0;
      while (y % i == 0) y /= i, cnt++;
      ret.emplace_back(i, cnt);
    }
  }
  if (y != 1) ret.emplace_back(y, 1);
  return ret;
}
void solve() {
  ll p, q;
  cin >> p >> q;
  if (p % q) {
    cout << p << '\n';
    return;
  }
  auto get = gogo(q);
  ll ans = 0;
  for (int i = 0; i < get.size(); i++) {
    ll su1 = power(get[i].first, get[i].second);
    ll t = p;
    while (t % get[i].first == 0) t /= get[i].first;
    t *= su1 / get[i].first;
    ans = max(ans, t);
  }
  cout << ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  int t;
  cin >> t;
  while (t--) solve();
}

```