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.

from math import gcd
def prime_decomposition(n):
    i = 2
    table = dict()
    while i * i <= n:
        c = 0
        while n % i == 0:
            n /= i
            c+=1
        if c!=0:
            table[i] = c
        i += 1
    if n > 1:
        table[int(n)] = 1
    return table

def f(N, mn, ds):
    # N以下mn以上でdsのどれでも割れない数
    M = len(ds)
    r = 0
    for i in range(1, 2**M):
        t,l=0,1
        for j in range(M):
            if i % 2:
                l*=ds[j]
                t += 1
            i = i>>1
        # N//l : N以下のlの倍数の数
        # (mn-1)//l : mn-1以下のlの倍数の数
        r += (1 if t%2 else -1)*(N//l - (mn-1)//l)
    return r

t = int(input())
for _ in range(t):
    a, m = map(int, input().split())
    aa = prime_decomposition(a)
    mm = prime_decomposition(m)
    ds = list()
    g = gcd(a, m)
    for key in mm:
        if key in aa and aa[key] >= mm[key]:
            continue
        ds.append(key)
    # a+xがdの倍数だとだめ
    mx = (a+m)//g-1
    mn = a//g
    print(mx-mn-f(mx, mn, ds)+1)
