You are given two integers a and m. Calculate the number of integers x such that 0 ≤ x < m and \gcd(a, m) = \gcd(a + x, m).

Note: \gcd(a, b) is the greatest common divisor of a and b.

Input

The first line contains the single integer T (1 ≤ T ≤ 50) — the number of test cases.

Next T lines contain test cases — one per line. Each line contains two integers a and m (1 ≤ a < m ≤ 10^{10}).

Output

Print T integers — one per test case. For each test case print the number of appropriate x-s.

Example

Input


3
4 9
5 10
42 9999999967


Output


6
1
9999999966

Note

In the first test case appropriate x-s are [0, 1, 3, 4, 6, 7].

In the second test case the only appropriate x is 0.
#include <bits/stdc++.h>
#pragma warning(disable : 4996)
#pragma comment(linker, "/STACK:16777216")
using namespace std;
const int INF = 2000000000;
const long long LINF = 2000000000000000000;
template <typename T>
void print(vector<T>& a) {
  for (int i = 0; i < a.size(); i++) cout << a[i] << ' ';
  cout << '\n';
}
template <typename T>
void print(deque<T>& a) {
  for (int i = 0; i < a.size(); i++) cout << a[i] << ' ';
  cout << '\n';
}
template <typename T>
void print(vector<vector<T>>& a) {
  for (int i = 0; i < a.size(); i++) {
    for (int j = 0; j < a[i].size(); j++) cout << a[i][j] << ' ';
    cout << '\n';
  }
}
template <typename T>
void input(vector<T>& a) {
  for (int i = 0; i < a.size(); i++) cin >> a[i];
}
template <typename T>
void input(deque<T>& a) {
  for (int i = 0; i < a.size(); i++) cin >> a[i];
}
template <typename T>
void input(vector<vector<T>>& a) {
  for (int i = 0; i < a.size(); i++)
    for (int j = 0; j < a[i].size(); j++) cin >> a[i][j];
}
long long gcd(long long a, long long b) {
  while (b) {
    a %= b;
    swap(a, b);
  }
  return a;
}
void solve() {
  long long a, m;
  cin >> a >> m;
  long long r = a + m - 1;
  long long l = a - 1;
  long long k = gcd(a, m);
  long long mm = m;
  mm /= k;
  vector<long long> factors;
  for (long long i = 2; i * i <= mm; i++) {
    if (mm % i == 0) {
      factors.push_back(i);
      while (mm % i == 0) mm /= i;
    }
  }
  if (mm != 1) factors.push_back(mm);
  sort((factors).begin(), (factors).end());
  int n = factors.size();
  int N = (1 << n);
  long long ans = r / k - l / k;
  for (int mask = 1; mask < N; mask++) {
    long long temp = 1;
    int cnt = 0;
    for (int i = 0; i < n; i++)
      if (mask & (1 << i)) {
        cnt++;
        temp *= factors[i];
      }
    temp *= k;
    if (cnt % 2)
      ans = ans - (r / temp - l / temp);
    else
      ans = ans + (r / temp - l / temp);
  }
  cout << ans << '\n';
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int tst = 1;
  cin >> tst;
  while (tst--) solve();
}
