#include <bits/stdc++.h>
#pragma GCC optimize("unroll-loops,no-stack-protector")
#pragma GCC target("sse,sse2,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
long long gcd(long long a, long long b) { return (!b ? a : gcd(b, a % b)); }
long long lcm(long long a, long long b) { return ((a * b) / gcd(a, b)); }
const long long MOD = 998244353;
const int INF32 = 1 << 30;
const long long INF64 = 1LL << 60;
const long double pi = 3.141592653589793;
long long modpow(long long n, long long k, long long mod) {
  if (k == 0) return 1;
  long long r = modpow(n * n % mod, k >> 1, mod);
  if (k & 1) r = r * n % mod;
  return r;
}
void solve() {
  int t;
  cin >> t;
  while (t--) {
    long long p, q;
    cin >> p >> q;
    if (p % q) {
      cout << p << endl;
      continue;
    }
    vector<long long> num;
    long long ans = p;
    for (long long i = 1; i <= sqrt(q); i++) {
      if (q % i == 0) {
        num.push_back(i);
        num.push_back(q / i);
      }
    }
    for (long long i = 0; i < num.size(); i++) {
      if (num[i] == 1) continue;
      long long pp = p;
      long long cnt = 0;
      while (pp % num[i] == 0) {
        if (pp % q) break;
        pp /= num[i];
        cnt++;
      }
      long long tmp = pow(num[i], cnt);
      ans = min(ans, tmp);
    }
    cout << p / ans << endl;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  solve();
  return 0;
}

### Prompt

Create a solution in cpp for 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>
#pragma GCC optimize("unroll-loops,no-stack-protector")
#pragma GCC target("sse,sse2,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
long long gcd(long long a, long long b) { return (!b ? a : gcd(b, a % b)); }
long long lcm(long long a, long long b) { return ((a * b) / gcd(a, b)); }
const long long MOD = 998244353;
const int INF32 = 1 << 30;
const long long INF64 = 1LL << 60;
const long double pi = 3.141592653589793;
long long modpow(long long n, long long k, long long mod) {
  if (k == 0) return 1;
  long long r = modpow(n * n % mod, k >> 1, mod);
  if (k & 1) r = r * n % mod;
  return r;
}
void solve() {
  int t;
  cin >> t;
  while (t--) {
    long long p, q;
    cin >> p >> q;
    if (p % q) {
      cout << p << endl;
      continue;
    }
    vector<long long> num;
    long long ans = p;
    for (long long i = 1; i <= sqrt(q); i++) {
      if (q % i == 0) {
        num.push_back(i);
        num.push_back(q / i);
      }
    }
    for (long long i = 0; i < num.size(); i++) {
      if (num[i] == 1) continue;
      long long pp = p;
      long long cnt = 0;
      while (pp % num[i] == 0) {
        if (pp % q) break;
        pp /= num[i];
        cnt++;
      }
      long long tmp = pow(num[i], cnt);
      ans = min(ans, tmp);
    }
    cout << p / ans << endl;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  solve();
  return 0;
}

```