#include <bits/stdc++.h>
using namespace std;
const double pi = acos(-1.0);
const double eps = 1e-6;
const int mod = 1e9 + 7;
const int INF = 0x3f3f3f3f;
int t;
long long p, q;
vector<pair<int, int> > ve;
long long qpow(long long a, long long p) {
  long long res = 1;
  while (p) {
    if (p & 1) res = res * a;
    a = a * a;
    p >>= 1;
  }
  return res;
}
int main(void) {
  scanf("%d", &t);
  while (t--) {
    scanf("%lld%lld", &p, &q);
    if (p < q) {
      printf("%lld\n", p);
    } else {
      if (p % q != 0) {
        printf("%lld\n", p);
      } else {
        ve.clear();
        for (int i = 2; 1ll * i * i <= q; i++) {
          if (q % i == 0) {
            int cnt = 0;
            while (q % i == 0) q /= i, ++cnt;
            ve.push_back(make_pair(i, cnt));
          }
        }
        if (q != 1) {
          ve.push_back(make_pair(q, 1));
        }
        long long mn = (long long)1e18;
        for (auto x : ve) {
          int v = x.first, c = x.second;
          long long tmp = p;
          int cnt = 0;
          while (tmp % v == 0) {
            ++cnt;
            tmp /= v;
          }
          long long tt = qpow(v, cnt - c + 1);
          mn = min(mn, tt);
        }
        printf("%lld\n", p / mn);
      }
    }
  }
  return 0;
}

### Prompt

Your challenge is to write a CPP solution to 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 double pi = acos(-1.0);
const double eps = 1e-6;
const int mod = 1e9 + 7;
const int INF = 0x3f3f3f3f;
int t;
long long p, q;
vector<pair<int, int> > ve;
long long qpow(long long a, long long p) {
  long long res = 1;
  while (p) {
    if (p & 1) res = res * a;
    a = a * a;
    p >>= 1;
  }
  return res;
}
int main(void) {
  scanf("%d", &t);
  while (t--) {
    scanf("%lld%lld", &p, &q);
    if (p < q) {
      printf("%lld\n", p);
    } else {
      if (p % q != 0) {
        printf("%lld\n", p);
      } else {
        ve.clear();
        for (int i = 2; 1ll * i * i <= q; i++) {
          if (q % i == 0) {
            int cnt = 0;
            while (q % i == 0) q /= i, ++cnt;
            ve.push_back(make_pair(i, cnt));
          }
        }
        if (q != 1) {
          ve.push_back(make_pair(q, 1));
        }
        long long mn = (long long)1e18;
        for (auto x : ve) {
          int v = x.first, c = x.second;
          long long tmp = p;
          int cnt = 0;
          while (tmp % v == 0) {
            ++cnt;
            tmp /= v;
          }
          long long tt = qpow(v, cnt - c + 1);
          mn = min(mn, tt);
        }
        printf("%lld\n", p / mn);
      }
    }
  }
  return 0;
}

```