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 long long maxn = 1e6 + 10;
const long long maxM = 1e6 + 10;
const long long inf = 1e8;
const long long inf2 = 1e17;
template <class T>
void read(T& x) {
  T s = 0, w = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') s = s * 10 + ch - '0', ch = getchar();
  x = s * w;
}
template <class H, class... T>
void read(H& h, T&... t) {
  read(h);
  read(t...);
}
void pt() { cout << '\n'; }
template <class H, class... T>
void pt(H h, T... t) {
  cout << " " << h;
  pt(t...);
}
int T;
long long a, b;
int P[maxn], tail;
bool vis[maxn];
map<int, int> mp;
long long A, B;
void initP(int N) {
  vis[1] = 1;
  for (int i = 2; i <= N; i++) {
    if (!vis[i]) P[tail++] = i;
    for (int j = 0; j < tail && P[j] <= N / i; j++) {
      vis[P[j] * i] = 1;
      if (i % P[j] == 0) break;
    }
  }
}
long long ksm(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = res * a;
    b >>= 1;
    a *= a;
  }
  return res;
}
void solve() {
  read(A, B);
  long long a = A, b = B;
  mp.clear();
  for (int i = 0; i < tail && P[i] * P[i] <= b; i++) {
    if (b % P[i] == 0) {
      int cnt = 0;
      while (b % P[i] == 0) {
        cnt++;
        b /= P[i];
      }
      mp[P[i]] = cnt;
    }
  }
  if (b > 1) mp[b] = 1;
  int find = 0;
  for (auto pp : mp) {
    long long p = pp.first, cnt = pp.second;
    long long coun = 0, a = A;
    while (a % p == 0) {
      coun++;
      a /= p;
    }
    if (coun < cnt) {
      find = 1;
      break;
    }
  }
  if (find) {
    printf("%lld\n", A);
  } else {
    long long ans = 0;
    for (auto pp : mp) {
      long long p = pp.first, cnt = pp.second;
      long long coun = 0, a = A;
      while (a % p == 0) {
        coun++;
        a /= p;
      }
      if (coun >= cnt) {
        ans = max(ans, A / ksm(p, coun - cnt + 1));
      }
    }
    printf("%lld\n", ans);
  }
}
int main() {
  initP(1000000);
  read(T);
  while (T--) {
    solve();
  }
  return 0;
}
