#include <bits/stdc++.h>
using namespace std;
using ld = long double;
using ll = long long;
const char nl = '\n';
const char sp = ' ';
const long long mod1 = 1e9 + 7;
const long long mod2 = 998244353;
const ld PI = acos(-1);
template <class c>
struct rge {
  c b, e;
};
template <class c>
rge<c> range(c i, c j) {
  return rge<c>{i, j};
}
template <class c>
auto dud(c* x) -> decltype(cerr << *x, 0);
template <class c>
char dud(...);
struct debug {
  template <class c>
  debug& operator<<(const c&) {
    return *this;
  }
};
class A {
 public:
  long long pwr(long long a, long long b) {
    if (a == 1 or b == 0) {
      return 1ll;
    }
    if (b == 1) {
      return a;
    }
    long long p = pwr(a, b / 2);
    p *= p;
    if (b & 1) p *= a;
    return p;
  }
  A() {
    long long t = 1;
    cin >> t;
    while (t--) {
      solve();
    }
  }
  void solve() {
    long long p, q;
    cin >> p >> q;
    if (p % q != 0) {
      cout << p << nl;
      return;
    }
    debug() << p << q;
    unordered_map<long long, long long> pQ;
    long long Q = q;
    while (q % 2 == 0) {
      q /= 2;
      pQ[2]++;
    }
    for (long long i = 3; i * i <= q; ++i) {
      while (q % i == 0) {
        pQ[i]++;
        q /= i;
      }
    }
    if (q != 1) {
      pQ[q]++;
    }
    q = Q;
    vector<long long> facs;
    debug() << p << q;
    unordered_map<long long, long long> pP;
    for (auto x : pQ) {
      long long a = x.first;
      facs.emplace_back(a);
      long long P = p;
      while (p % a == 0) {
        pP[a]++;
        p /= a;
      }
      p = P;
    }
    long long ans = -1;
    for (auto x : facs) {
      long long z = pP[x] - pQ[x] + 1;
      if (ans == -1) {
        ans = pwr(x, z);
      } else {
        ans = min(ans, pwr(x, z));
      }
    }
    debug() << p << q << ans;
    ans = p / ans;
    cout << ans << nl;
  }
};
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  A a;
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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;
using ld = long double;
using ll = long long;
const char nl = '\n';
const char sp = ' ';
const long long mod1 = 1e9 + 7;
const long long mod2 = 998244353;
const ld PI = acos(-1);
template <class c>
struct rge {
  c b, e;
};
template <class c>
rge<c> range(c i, c j) {
  return rge<c>{i, j};
}
template <class c>
auto dud(c* x) -> decltype(cerr << *x, 0);
template <class c>
char dud(...);
struct debug {
  template <class c>
  debug& operator<<(const c&) {
    return *this;
  }
};
class A {
 public:
  long long pwr(long long a, long long b) {
    if (a == 1 or b == 0) {
      return 1ll;
    }
    if (b == 1) {
      return a;
    }
    long long p = pwr(a, b / 2);
    p *= p;
    if (b & 1) p *= a;
    return p;
  }
  A() {
    long long t = 1;
    cin >> t;
    while (t--) {
      solve();
    }
  }
  void solve() {
    long long p, q;
    cin >> p >> q;
    if (p % q != 0) {
      cout << p << nl;
      return;
    }
    debug() << p << q;
    unordered_map<long long, long long> pQ;
    long long Q = q;
    while (q % 2 == 0) {
      q /= 2;
      pQ[2]++;
    }
    for (long long i = 3; i * i <= q; ++i) {
      while (q % i == 0) {
        pQ[i]++;
        q /= i;
      }
    }
    if (q != 1) {
      pQ[q]++;
    }
    q = Q;
    vector<long long> facs;
    debug() << p << q;
    unordered_map<long long, long long> pP;
    for (auto x : pQ) {
      long long a = x.first;
      facs.emplace_back(a);
      long long P = p;
      while (p % a == 0) {
        pP[a]++;
        p /= a;
      }
      p = P;
    }
    long long ans = -1;
    for (auto x : facs) {
      long long z = pP[x] - pQ[x] + 1;
      if (ans == -1) {
        ans = pwr(x, z);
      } else {
        ans = min(ans, pwr(x, z));
      }
    }
    debug() << p << q << ans;
    ans = p / ans;
    cout << ans << nl;
  }
};
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  A a;
  return 0;
}

```