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 = 1e5 + 100;
const long long mod = 998244353, inf = 1e18;
const long long g = 3;
template <class T>
void Mul(T &x, T y) {
  x = x * y % mod;
}
template <class T>
void Add(T &x, T y) {
  x = (x + y) % mod;
}
template <class T>
void Sub(T &x, T y) {
  x = (x - y + mod) % mod;
}
template <class T>
void Mn(T &x, T y) {
  x = min(x, y);
}
template <class T>
void Mx(T &x, T y) {
  x = max(x, y);
}
long long qpow(long long x, long long y = mod - 2) {
  long long res = 1;
  while (y) {
    if (y & 1) Mul(res, x);
    Mul(x, x), y >>= 1;
  }
  return res;
}
void Div(long long &x, long long y) { Mul(x, qpow(y)); }
int n, m, k, T;
long long gcd(long long x, long long y) { return !y ? x : gcd(y, x % y); }
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> T;
  long long p, q, ans, tmp;
  while (T--) {
    cin >> p >> q;
    if (p < q) {
      cout << p << '\n';
      continue;
    }
    if (p % q != 0) {
      cout << p << '\n';
      continue;
    }
    ans = p;
    while (ans % q == 0) ans /= q;
    for (long long i = 2, j, t; i * i <= q; i++) {
      if (q % i == 0) {
        tmp = p;
        while (tmp % q == 0) tmp /= i;
        ans = max(ans, tmp);
        j = q / i;
        tmp = p;
        while (tmp % q == 0) tmp /= j;
        ans = max(ans, tmp);
      }
    }
    cout << ans << '\n';
  }
}
