#include <bits/stdc++.h>
using namespace std;
const int dx[4] = {-1, 0, 1, 0}, dy[4] = {0, 1, 0, -1};
const int dxx[8] = {-1, -1, 0, 1, 1, 1, 0, -1},
          dyy[8] = {0, 1, 1, 1, 0, -1, -1, -1};
const long long mod = 1000000007;
const int base = 311;
vector<pair<int, int> > fact(long long k) {
  vector<pair<int, int> > ans;
  for (int i = 2; i * i <= k; ++i) {
    if (k % i == 0) {
      int dem = 0;
      while (k % i == 0) {
        k /= i;
        ++dem;
      }
      ans.push_back(make_pair(i, dem));
    }
  }
  if (k > 1) ans.push_back(make_pair(k, 1));
  return ans;
}
long long pw(long long n, long long k) {
  if (k == 0) return 1;
  long long t = pw(n, k / 2);
  if (k & 1)
    return t * t * n;
  else
    return t * t;
}
void gogo() {
  long long p, q;
  cin >> p >> q;
  vector<pair<int, int> > v1 = fact(q);
  vector<pair<int, int> > v2;
  long long ans = 0;
  for (pair<int, int> i : v1) {
    long long k = p, cnt = 0;
    while (k % i.first == 0) {
      k /= i.first;
      ++cnt;
    }
    if (cnt < i.second)
      ans = p;
    else {
      ans = max(ans, p / pw(i.first, (cnt - i.second + 1)));
    }
  }
  cout << ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  if (fopen("sol"
            ".inp",
            "r")) {
    freopen(
        "sol"
        ".inp",
        "r", stdin);
    freopen(
        "sol"
        ".out",
        "w", stdout);
  }
  int t;
  cin >> t;
  while (t--) gogo();
  return 0;
}

### Prompt

Your challenge is to write 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;
const int dx[4] = {-1, 0, 1, 0}, dy[4] = {0, 1, 0, -1};
const int dxx[8] = {-1, -1, 0, 1, 1, 1, 0, -1},
          dyy[8] = {0, 1, 1, 1, 0, -1, -1, -1};
const long long mod = 1000000007;
const int base = 311;
vector<pair<int, int> > fact(long long k) {
  vector<pair<int, int> > ans;
  for (int i = 2; i * i <= k; ++i) {
    if (k % i == 0) {
      int dem = 0;
      while (k % i == 0) {
        k /= i;
        ++dem;
      }
      ans.push_back(make_pair(i, dem));
    }
  }
  if (k > 1) ans.push_back(make_pair(k, 1));
  return ans;
}
long long pw(long long n, long long k) {
  if (k == 0) return 1;
  long long t = pw(n, k / 2);
  if (k & 1)
    return t * t * n;
  else
    return t * t;
}
void gogo() {
  long long p, q;
  cin >> p >> q;
  vector<pair<int, int> > v1 = fact(q);
  vector<pair<int, int> > v2;
  long long ans = 0;
  for (pair<int, int> i : v1) {
    long long k = p, cnt = 0;
    while (k % i.first == 0) {
      k /= i.first;
      ++cnt;
    }
    if (cnt < i.second)
      ans = p;
    else {
      ans = max(ans, p / pw(i.first, (cnt - i.second + 1)));
    }
  }
  cout << ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  if (fopen("sol"
            ".inp",
            "r")) {
    freopen(
        "sol"
        ".inp",
        "r", stdin);
    freopen(
        "sol"
        ".out",
        "w", stdout);
  }
  int t;
  cin >> t;
  while (t--) gogo();
  return 0;
}

```