#include <bits/stdc++.h>
using namespace std;
vector<pair<long long, long long> > lst;
long long mul, ans, p, q, q2;
void solve(long long pos, long long val) {
  if (pos == lst.size()) {
    const long long mod = q2;
    if (val % mod != 0) ans = max(ans, val);
    return;
  }
  long long num = 1;
  for (long long i = 0; i <= lst[pos].second; i++) {
    solve(pos + 1, val * num);
    num *= lst[pos].first;
  }
}
int main() {
  long long t;
  cin >> t;
  while (t--) {
    scanf("%lld", &p), scanf("%lld", &q);
    q2 = q;
    if (p % q != 0) {
      printf("%lld\n", p);
      continue;
    }
    mul = p;
    for (long long i = 2; i * i <= q; i++) {
      if (q % i == 0) {
        while (q % i == 0) {
          q /= i;
        }
        long long cnt = 0;
        while (mul % i == 0) {
          mul /= i;
          cnt++;
        }
        if (cnt != 0) lst.push_back(make_pair(i, cnt));
      }
    }
    if (q > 1 && mul % q == 0) {
      long long cnt = 0;
      while (mul % q == 0) {
        mul /= q;
        cnt++;
      }
      lst.push_back(make_pair(q, cnt));
    }
    ans = mul;
    solve(0, mul);
    printf("%lld\n", ans);
    lst.clear();
  }
  return 0;
}

### 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;
vector<pair<long long, long long> > lst;
long long mul, ans, p, q, q2;
void solve(long long pos, long long val) {
  if (pos == lst.size()) {
    const long long mod = q2;
    if (val % mod != 0) ans = max(ans, val);
    return;
  }
  long long num = 1;
  for (long long i = 0; i <= lst[pos].second; i++) {
    solve(pos + 1, val * num);
    num *= lst[pos].first;
  }
}
int main() {
  long long t;
  cin >> t;
  while (t--) {
    scanf("%lld", &p), scanf("%lld", &q);
    q2 = q;
    if (p % q != 0) {
      printf("%lld\n", p);
      continue;
    }
    mul = p;
    for (long long i = 2; i * i <= q; i++) {
      if (q % i == 0) {
        while (q % i == 0) {
          q /= i;
        }
        long long cnt = 0;
        while (mul % i == 0) {
          mul /= i;
          cnt++;
        }
        if (cnt != 0) lst.push_back(make_pair(i, cnt));
      }
    }
    if (q > 1 && mul % q == 0) {
      long long cnt = 0;
      while (mul % q == 0) {
        mul /= q;
        cnt++;
      }
      lst.push_back(make_pair(q, cnt));
    }
    ans = mul;
    solve(0, mul);
    printf("%lld\n", ans);
    lst.clear();
  }
  return 0;
}

```