#include <bits/stdc++.h>
using namespace std;
const long long maxn = 1e5 + 10;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long prime[maxn], vis[maxn], cnt;
void init() {
  for (long long i = 2; i < maxn; i++) {
    if (!vis[i]) {
      prime[cnt++] = i;
      for (long long j = 1; j * i < maxn; j++) vis[i * j] = 1;
    }
  }
}
long long a[maxn], num[maxn], sum;
void fj(long long x) {
  sum = 0;
  long long i = 0;
  while (i < cnt && prime[i] <= x) {
    long long k = prime[i];
    if (x % k == 0) {
      long long s = 0;
      while (x % k == 0) x /= k, s++;
      a[sum] = k;
      num[sum++] = s;
    }
    i++;
  }
  if (x > 1) a[sum] = x, num[sum++] = 1;
}
long long solve(long long x) {
  long long ss = 1e18 + 10;
  for (long long i = 0; i < sum; i++) {
    long long s = 0;
    long long k = 1;
    while (x % a[i] == 0) {
      x /= a[i];
      s++;
      if (s >= num[i]) k *= a[i];
    }
    ss = min(k, ss);
  }
  return ss;
}
signed main() {
  long long n;
  cin >> n;
  init();
  while (n--) {
    long long a, b;
    cin >> a >> b;
    if (a % b) {
      cout << a << "\n";
      continue;
    }
    fj(b);
    cout << a / solve(a) << "\n";
  }
  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 maxn = 1e5 + 10;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long prime[maxn], vis[maxn], cnt;
void init() {
  for (long long i = 2; i < maxn; i++) {
    if (!vis[i]) {
      prime[cnt++] = i;
      for (long long j = 1; j * i < maxn; j++) vis[i * j] = 1;
    }
  }
}
long long a[maxn], num[maxn], sum;
void fj(long long x) {
  sum = 0;
  long long i = 0;
  while (i < cnt && prime[i] <= x) {
    long long k = prime[i];
    if (x % k == 0) {
      long long s = 0;
      while (x % k == 0) x /= k, s++;
      a[sum] = k;
      num[sum++] = s;
    }
    i++;
  }
  if (x > 1) a[sum] = x, num[sum++] = 1;
}
long long solve(long long x) {
  long long ss = 1e18 + 10;
  for (long long i = 0; i < sum; i++) {
    long long s = 0;
    long long k = 1;
    while (x % a[i] == 0) {
      x /= a[i];
      s++;
      if (s >= num[i]) k *= a[i];
    }
    ss = min(k, ss);
  }
  return ss;
}
signed main() {
  long long n;
  cin >> n;
  init();
  while (n--) {
    long long a, b;
    cin >> a >> b;
    if (a % b) {
      cout << a << "\n";
      continue;
    }
    fj(b);
    cout << a / solve(a) << "\n";
  }
  return 0;
}

```