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 math

T=int(input())
for _ in range(T):
    a,m=map(int,input().split())

    t=math.gcd(a,m)

    ans=(m-1)//t
    print(ans)




import math

T=int(input())
for _ in range(T):
    a,m=map(int,input().split())

    ans=0
    t=math.gcd(a,m)
    print('trg',t)
    for i in range(0,m):
        #print(i+a,m)
        if(math.gcd(i+a,m)==t):
            print(i+a,m)
            ans+=1
    print(ans)
"""


import math

def lcm(a,b):
    ans=(a*b)//math.gcd(a,b)
    return ans

T=int(input())
for _ in range(T):
    a,m=map(int,input().split())

    t=math.gcd(a,m)

    ans=m//t
    fnl=m//t

    for i in range(2, int(math.sqrt(m) + 1)) :

        if(ans%i==0):
            while(ans%i==0):
                ans=ans//i
            fnl-=(fnl//i)

    if(ans>1):
        fnl-=(fnl//ans)
        

    print(fnl)
            

            
                  
    

    

    #ans=(m-1)//t
    #print(ans)

