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.
import java.util.*;
import java.io.*;
 
public class Solution{
	static PrintWriter out=new PrintWriter(System.out);
	public static void main (String[] args) throws IOException{
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		
		String[] input=br.readLine().trim().split(" ");
		
		int numTestCases=Integer.parseInt(input[0]);
		while(numTestCases-->0){
		    input=br.readLine().trim().split(" ");
    		long a=Long.parseLong(input[0]);
    		long m=Long.parseLong(input[1]);
    		
    		long gcd=gcd(a,m);
    		long ans=eulerTotient(m/gcd);
    		out.println(ans);
		}
		
		out.flush();
		out.close();
	}	
	
	public static long gcd(long a,long b)
	{
	    if(a==0){
	        return b;
	    }
	    if(b==0){
	        return a;
	    }
	    return gcd(b%a,a);
	}
	
	public static long eulerTotient(long n)
	{
	    long result = n;
        for (int i = 2; 1L * i * i <= n; i++) {
            if (n % i == 0) {
                while (n % i == 0)
                    n /= i;
                result -= result / i;
            }
        }
        if (n > 1)
            result -= result / n;
        return result;
	}
	
	
	
}