#include <bits/stdc++.h>
using namespace std;
long long int pow1(long long int n, long long int k) {
  if (k == 0) return 1;
  if (k == 1) return n;
  long long int X = 1;
  if (k % 2 == 1) X = n;
  long long int Y = pow1(n, k / 2);
  return (((Y * Y)) * X);
}
const long long int maxN = 100001;
map<long long int, long long int> Vis;
vector<long long int> V[maxN];
void dfs(long long int node) {
  if (Vis[node]) return;
  Vis[node] = 1;
  for (auto child : V[node]) {
    dfs(child);
  }
  return;
}
void setIO(string s = "") {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin.exceptions(cin.failbit);
  if (s.size()) {
    freopen((s + ".in").c_str(), "r", stdin);
    freopen((s + ".out").c_str(), "w", stdout);
  }
}
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
void solve() {
  long long int n, m, j, k, l, i;
  cin >> n >> m;
  long long int ans = 1;
  long long int g = gcd(n, m);
  long long int N = n / g;
  long long int M = m / g;
  if (gcd(n, m) != m) {
    ans = n;
  } else {
    long long int c1 = 0;
    long long int H = m;
    i = m;
    while (H % i == 0) {
      H /= i;
      c1++;
    }
    long long int K = n;
    while (K % i == 0) {
      K /= i;
    }
    K *= pow1(i, c1 - 1);
    ans = max(ans, K);
    for (i = 2; i * i <= m; i++) {
      if (m % i == 0) {
        long long int c1 = 0;
        long long int H = m;
        while (H % i == 0) {
          H /= i;
          c1++;
        }
        long long int K = n;
        while (K % i == 0) {
          K /= i;
        }
        K *= pow1(i, c1 - 1);
        ans = max(ans, K);
        c1 = 0;
        H = m;
        while (H % (m / i) == 0) {
          H /= (m / i);
          c1++;
        }
        K = n;
        while (K % (m / i) == 0) {
          K /= (m / i);
        }
        K *= pow1((m / i), c1 - 1);
        ans = max(ans, K);
      }
    }
  }
  cout << ans;
  cout << "\n";
  return;
}
int main() {
  setIO();
  long long int totalTestCase = 1;
  cin >> totalTestCase;
  for (long long int testCase = 1; testCase <= totalTestCase; testCase++) {
    solve();
  }
  return 0;
}

### Prompt

Generate 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;
long long int pow1(long long int n, long long int k) {
  if (k == 0) return 1;
  if (k == 1) return n;
  long long int X = 1;
  if (k % 2 == 1) X = n;
  long long int Y = pow1(n, k / 2);
  return (((Y * Y)) * X);
}
const long long int maxN = 100001;
map<long long int, long long int> Vis;
vector<long long int> V[maxN];
void dfs(long long int node) {
  if (Vis[node]) return;
  Vis[node] = 1;
  for (auto child : V[node]) {
    dfs(child);
  }
  return;
}
void setIO(string s = "") {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin.exceptions(cin.failbit);
  if (s.size()) {
    freopen((s + ".in").c_str(), "r", stdin);
    freopen((s + ".out").c_str(), "w", stdout);
  }
}
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
void solve() {
  long long int n, m, j, k, l, i;
  cin >> n >> m;
  long long int ans = 1;
  long long int g = gcd(n, m);
  long long int N = n / g;
  long long int M = m / g;
  if (gcd(n, m) != m) {
    ans = n;
  } else {
    long long int c1 = 0;
    long long int H = m;
    i = m;
    while (H % i == 0) {
      H /= i;
      c1++;
    }
    long long int K = n;
    while (K % i == 0) {
      K /= i;
    }
    K *= pow1(i, c1 - 1);
    ans = max(ans, K);
    for (i = 2; i * i <= m; i++) {
      if (m % i == 0) {
        long long int c1 = 0;
        long long int H = m;
        while (H % i == 0) {
          H /= i;
          c1++;
        }
        long long int K = n;
        while (K % i == 0) {
          K /= i;
        }
        K *= pow1(i, c1 - 1);
        ans = max(ans, K);
        c1 = 0;
        H = m;
        while (H % (m / i) == 0) {
          H /= (m / i);
          c1++;
        }
        K = n;
        while (K % (m / i) == 0) {
          K /= (m / i);
        }
        K *= pow1((m / i), c1 - 1);
        ans = max(ans, K);
      }
    }
  }
  cout << ans;
  cout << "\n";
  return;
}
int main() {
  setIO();
  long long int totalTestCase = 1;
  cin >> totalTestCase;
  for (long long int testCase = 1; testCase <= totalTestCase; testCase++) {
    solve();
  }
  return 0;
}

```