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.
#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 10, M = 2e6 + 10, inf = 0x3f3f3f3f;
const long long Linf = 0x3f3f3f3f3f3f3f3f;
inline int read() {
  bool sym = 0;
  int res = 0;
  char ch = getchar();
  while (!isdigit(ch)) sym |= (ch == '-'), ch = getchar();
  while (isdigit(ch)) res = (res << 3) + (res << 1) + (ch ^ 48), ch = getchar();
  return sym ? -res : res;
}
inline long long Lread() {
  bool sym = 0;
  long long res = 0;
  char ch = getchar();
  while (!isdigit(ch)) sym |= (ch == '-'), ch = getchar();
  while (isdigit(ch)) res = (res << 3) + (res << 1) + (ch ^ 48), ch = getchar();
  return sym ? -res : res;
}
int n, m, cnt;
long long a, b, prime[N];
bool flag[N];
void init(long long N) {
  flag[1] = 0;
  for (long long i = 2; i <= N; i++) {
    if (!flag[i]) prime[++cnt] = i;
    for (int j = 1; j <= cnt && i * prime[j] <= N; j++) {
      flag[i * prime[j]] = 1;
      if (i % prime[j] == 0) break;
    }
  }
}
int main() {
  int T = read();
  init(100000);
  while (T--) {
    a = Lread();
    b = Lread();
    long long ans = Linf, c = a;
    if (a % b != 0) {
      printf("%lld\n", a);
      continue;
    }
    for (int i = 1; i <= cnt; i++) {
      if (a % prime[i] == 0) {
        long long ta = 1, tb = 1;
        while (a % prime[i] == 0) a /= prime[i], ta *= prime[i];
        while (b % prime[i] == 0) b /= prime[i], tb *= prime[i];
        if (tb == 1) continue;
        if (ta >= tb) ans = min(ans, ta / tb * prime[i]);
      }
    }
    if (b != 1) {
      long long ta = 1;
      while (a % b == 0) a /= b, ta *= b;
      ans = min(ans, ta);
    }
    printf("%lld\n", c / ans);
  }
  return 0;
}
