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;
template <typename T>
void view(T e) {
  std::cerr << e << std::endl;
}
template <typename T>
void view(const std::vector<T>& v) {
  for (const auto& e : v) {
    std::cerr << e << " ";
  }
  std::cerr << std::endl;
}
template <typename T>
void view(const std::vector<std::vector<T>>& vv) {
  cerr << endl;
  int cnt = 0;
  for (const auto& v : vv) {
    cerr << cnt << "th : ";
    view(v);
    cnt++;
  }
  cerr << endl;
}
using ll = long long;
const ll mod = 1000000007;
const int inf = 1001001001;
const ll INF = 1001001001001001001ll;
int dx[] = {1, 0, -1, 0};
int dy[] = {0, 1, 0, -1};
template <class T, class K>
bool chmax(T& a, const K b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T, class K>
bool chmin(T& a, const K b) {
  if (b < a) {
    a = b;
    return 1;
  }
  return 0;
}
ll rudiv(ll a, ll b) { return a / b + ((a ^ b) > 0 && a % b); }
ll rddiv(ll a, ll b) { return a / b - ((a ^ b) < 0 && a % b); }
ll power(ll a, ll p) {
  ll ret = 1;
  while (p) {
    if (p & 1) {
      ret = ret * a;
    }
    a = a * a;
    p >>= 1;
  }
  return ret;
}
ll modpow(ll a, ll p, ll m) {
  ll ret = 1;
  while (p) {
    if (p & 1) {
      ret = ret * a % m;
    }
    a = a * a % m;
    p >>= 1;
  }
  return ret;
}
int main() {
  cin.tie(nullptr);
  ios::sync_with_stdio(false);
  int t;
  cin >> t;
  while (t--) {
    ll p, q;
    cin >> p >> q;
    if (p % q != 0) {
      cout << p << endl;
      continue;
    }
    vector<pair<ll, int>> prime;
    ll tmp = q;
    for (ll i = 2; i * i <= q; i++) {
      int cnt = 0;
      while (tmp % i == 0) {
        cnt++;
        tmp /= i;
      }
      if (cnt) prime.push_back({i, cnt});
    }
    if (tmp != 1) prime.push_back({tmp, 1});
    ll ans = 1;
    for (auto pa : prime) {
      int cnt = 0;
      ll tmp = p;
      while (tmp % pa.first == 0) {
        cnt++;
        tmp /= pa.first;
      }
      chmax(ans, p / power(pa.first, cnt - (pa.second - 1)));
    }
    cout << ans << endl;
  }
}
