#include <bits/stdc++.h>
using namespace std;
vector<long long int> sieve;
vector<long long int> *Sieve(int N) {
  const long long int maxn = N;
  sieve.resize(maxn);
  for (long long int(i) = 0; (i) < (maxn); ++(i)) sieve[i] = 1;
  sieve[1] = 0;
  sieve[0] = 0;
  for (long long int(i) = 2; (i) <= (maxn); (i) += (1))
    if (1 == sieve[i])
      for (long long int j = i * i; j < maxn; j += i) sieve[j] = 0;
  vector<long long int> *primes = new vector<long long int>();
  primes->push_back(2);
  for (int i = 3; i < maxn; i += 2)
    if (sieve[i]) {
      primes->push_back(i);
    }
  return primes;
}
long long int extended_GCD(long long int a, long long int b, long long int &x,
                           long long int &y) {
  if (a == 0) {
    x = 0;
    y = 1;
    return b;
  }
  long long int x1, y1;
  long long int gcd = extended_GCD(b % a, a, x1, y1);
  x = y1 - (b / a) * x1;
  y = x1;
  return gcd;
}
long long int ncr(long long int n, long long int k) {
  long long int r = 1;
  if (k > n - k) k = n - k;
  for (long long int i = 0; i < k; i++) {
    r *= (n - i);
    r /= (i + 1);
  }
  return r;
}
long long int power(long long int a, long long int b,
                    long long int m = 1000000007) {
  long long int res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a) % m;
    a = (a * a) % m;
    b >>= 1;
  }
  return res;
}
long long int modinv(long long int a, long long int mod = 1000000007) {
  long long int x, y;
  extended_GCD(a, mod, x, y);
  if (x < 0) x += mod;
  return x;
}
bool prime(int n) {
  int c = 0;
  for (int i = 2; (i * i) <= n; i++) {
    if (n % i == 0) {
      c++;
      return false;
    }
  }
  if (c == 0) return true;
}
vector<int> v[200006];
int vis[200006];
int c;
void dfs(int n) {
  vis[n] = 1;
  c++;
  for (int i : v[n]) {
    if (!vis[i]) dfs(i);
  }
}
map<long long int, long long int> factors(long long int b) {
  map<long long int, long long int> m;
  for (long long int i = 2; i * i <= b; i++) {
    if (b % i == 0) {
      while (b % i == 0) {
        m[i]++;
        b /= i;
      }
    }
  }
  if (b > 1) m[b]++;
  return m;
}
void solve() {
  long long int a, b, c, d;
  cin >> a >> b;
  long long int ans;
  if (a % b)
    ans = a;
  else {
    map<long long int, long long int> f;
    f = factors(b);
    ans = 0;
    for (auto i : f) {
      long long int t = a;
      while (t % b == 0) {
        t /= i.first;
      }
      ans = max(ans, t);
    }
  }
  cout << ans << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t;
  cin >> t;
  while (t--) solve();
  return 0;
}

### Prompt

Generate 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<long long int> sieve;
vector<long long int> *Sieve(int N) {
  const long long int maxn = N;
  sieve.resize(maxn);
  for (long long int(i) = 0; (i) < (maxn); ++(i)) sieve[i] = 1;
  sieve[1] = 0;
  sieve[0] = 0;
  for (long long int(i) = 2; (i) <= (maxn); (i) += (1))
    if (1 == sieve[i])
      for (long long int j = i * i; j < maxn; j += i) sieve[j] = 0;
  vector<long long int> *primes = new vector<long long int>();
  primes->push_back(2);
  for (int i = 3; i < maxn; i += 2)
    if (sieve[i]) {
      primes->push_back(i);
    }
  return primes;
}
long long int extended_GCD(long long int a, long long int b, long long int &x,
                           long long int &y) {
  if (a == 0) {
    x = 0;
    y = 1;
    return b;
  }
  long long int x1, y1;
  long long int gcd = extended_GCD(b % a, a, x1, y1);
  x = y1 - (b / a) * x1;
  y = x1;
  return gcd;
}
long long int ncr(long long int n, long long int k) {
  long long int r = 1;
  if (k > n - k) k = n - k;
  for (long long int i = 0; i < k; i++) {
    r *= (n - i);
    r /= (i + 1);
  }
  return r;
}
long long int power(long long int a, long long int b,
                    long long int m = 1000000007) {
  long long int res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a) % m;
    a = (a * a) % m;
    b >>= 1;
  }
  return res;
}
long long int modinv(long long int a, long long int mod = 1000000007) {
  long long int x, y;
  extended_GCD(a, mod, x, y);
  if (x < 0) x += mod;
  return x;
}
bool prime(int n) {
  int c = 0;
  for (int i = 2; (i * i) <= n; i++) {
    if (n % i == 0) {
      c++;
      return false;
    }
  }
  if (c == 0) return true;
}
vector<int> v[200006];
int vis[200006];
int c;
void dfs(int n) {
  vis[n] = 1;
  c++;
  for (int i : v[n]) {
    if (!vis[i]) dfs(i);
  }
}
map<long long int, long long int> factors(long long int b) {
  map<long long int, long long int> m;
  for (long long int i = 2; i * i <= b; i++) {
    if (b % i == 0) {
      while (b % i == 0) {
        m[i]++;
        b /= i;
      }
    }
  }
  if (b > 1) m[b]++;
  return m;
}
void solve() {
  long long int a, b, c, d;
  cin >> a >> b;
  long long int ans;
  if (a % b)
    ans = a;
  else {
    map<long long int, long long int> f;
    f = factors(b);
    ans = 0;
    for (auto i : f) {
      long long int t = a;
      while (t % b == 0) {
        t /= i.first;
      }
      ans = max(ans, t);
    }
  }
  cout << ans << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t;
  cin >> t;
  while (t--) solve();
  return 0;
}

```