#include <bits/stdc++.h>
using namespace std;
const long long N = 1e6 + 5;
const long long MOD = 998244353;
long long gcd(long long a, long long b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
long long ceiling(long long a, long long b) {
  if (a % b != 0) {
    return (a / b) + 1;
  } else
    return (a / b);
}
long long ncr(long long n, long long r, long long p) {
  if (r > n - r) r = n - r;
  long long c[r + 1];
  memset(c, 0, sizeof(c));
  c[0] = 1;
  for (long long i = 1; i <= n; i++) {
    for (long long j = min(i, r); j > 0; j--) c[j] = (c[j] + c[j - 1]) % p;
  }
  return c[r];
}
bool prime[N];
long long ans[N];
void sieve() {
  for (long long i = 0; i < N; i++) prime[i] = true;
  prime[0] = prime[1] = false;
  for (long long i = 4; i < N; i += 2) prime[i] = false;
  for (long long i = 3; i * i < N; i += 2) {
    if (prime[i]) {
      for (long long j = i * i; j < N; j += 2 * i) prime[j] = false;
    }
  }
}
long long findmsb(long long x) {
  for (long long i = 31; i >= 0; i--) {
    if ((1 << i) & x) return i;
  }
  return 100;
}
void solve() {
  long long p, q;
  long long temp;
  cin >> p >> q, temp = q;
  if (p % q > 0)
    cout << p << "\n";
  else {
    long long ans = 1;
    set<long long> s;
    for (long long i = 2; i <= q / i; i++)
      while (q % i == 0) {
        s.insert(i);
        q /= i;
      }
    s.insert(q), s.erase(1);
    q = temp;
    for (auto &i : s) {
      long long x = p;
      while (x % q == 0 && x > 0) x /= i;
      ans = max(ans, x);
    }
    cout << ans << "\n";
  }
  return;
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long tc;
  cin >> tc;
  while (tc--) {
    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>
using namespace std;
const long long N = 1e6 + 5;
const long long MOD = 998244353;
long long gcd(long long a, long long b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
long long ceiling(long long a, long long b) {
  if (a % b != 0) {
    return (a / b) + 1;
  } else
    return (a / b);
}
long long ncr(long long n, long long r, long long p) {
  if (r > n - r) r = n - r;
  long long c[r + 1];
  memset(c, 0, sizeof(c));
  c[0] = 1;
  for (long long i = 1; i <= n; i++) {
    for (long long j = min(i, r); j > 0; j--) c[j] = (c[j] + c[j - 1]) % p;
  }
  return c[r];
}
bool prime[N];
long long ans[N];
void sieve() {
  for (long long i = 0; i < N; i++) prime[i] = true;
  prime[0] = prime[1] = false;
  for (long long i = 4; i < N; i += 2) prime[i] = false;
  for (long long i = 3; i * i < N; i += 2) {
    if (prime[i]) {
      for (long long j = i * i; j < N; j += 2 * i) prime[j] = false;
    }
  }
}
long long findmsb(long long x) {
  for (long long i = 31; i >= 0; i--) {
    if ((1 << i) & x) return i;
  }
  return 100;
}
void solve() {
  long long p, q;
  long long temp;
  cin >> p >> q, temp = q;
  if (p % q > 0)
    cout << p << "\n";
  else {
    long long ans = 1;
    set<long long> s;
    for (long long i = 2; i <= q / i; i++)
      while (q % i == 0) {
        s.insert(i);
        q /= i;
      }
    s.insert(q), s.erase(1);
    q = temp;
    for (auto &i : s) {
      long long x = p;
      while (x % q == 0 && x > 0) x /= i;
      ans = max(ans, x);
    }
    cout << ans << "\n";
  }
  return;
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long tc;
  cin >> tc;
  while (tc--) {
    solve();
  }
  return 0;
}

```