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;
long long spf[3000001];
long long fac[3000001];
void sieve() {
  spf[1] = 1;
  for (long long i = 2; i < 3000001; i++) spf[i] = i;
  for (long long i = 4; i < 3000001; i += 2) spf[i] = 2;
  for (long long i = 3; i * i < 3000001; i++) {
    if (spf[i] == i) {
      for (long long j = i * i; j < 3000001; j += i)
        if (spf[j] == j) spf[j] = i;
    }
  }
}
map<long long, long long> getfactor(long long a) {
  map<long long, long long> m;
  while (a > 1) {
    m[spf[a]]++;
    a /= spf[a];
  }
  return m;
}
long long power(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long inverse(long long a, long long p) { return power(a, p - 2, p); }
long long ncr(long long n, long long r, long long p) {
  if (r > n) return 0;
  if (r == 0) return 1;
  return (fac[n] * inverse(fac[r], p) % p * inverse(fac[n - r], p) % p) % p;
}
void solve() {
  long long p, q;
  cin >> p >> q;
  long long temp = q;
  vector<long long> v;
  if (p % q == 0) {
    for (long long i = 2; i <= sqrt(q); i++) {
      if (temp % i == 0) {
        v.push_back(i);
        while (temp % i == 0) temp /= i;
      }
      if (i % 2) i++;
    }
    if (temp > 1) v.push_back(temp);
    long long ans = 0;
    for (auto it : v) {
      long long gg = p, hh = q;
      long long cnt1 = 0, cnt2 = 0;
      while (gg % it == 0) {
        cnt1++;
        gg /= it;
      }
      while (hh % it == 0) {
        cnt2++;
        hh /= it;
      }
      ans = max(ans, gg * power(it, cnt2 - 1, 1e18));
    }
    cout << ans << '\n';
  } else
    cout << p << '\n';
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long test = 1;
  cin >> test;
  while (test--) {
    solve();
  }
}
