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 sys import stdin
from collections import deque
mod = 10**9 + 7
import sys
sys.setrecursionlimit(10**5)
from queue import PriorityQueue
# def rl():
#     return [int(w) for w in stdin.readline().split()]
from bisect import bisect_right
from bisect import bisect_left
from collections import defaultdict
from math import sqrt,factorial,gcd,log2,inf,ceil
# map(int,input().split())
# # l = list(map(int,input().split()))
# from itertools import permutations
import heapq
# input = lambda: sys.stdin.readline().rstrip()
input = lambda : sys.stdin.readline().rstrip()
from sys import stdin, stdout
from heapq import heapify, heappush, heappop


# n,k = map(int,input().split())
#
# l = list(map(int,input().split()))




t = int(input())

for _ in range(t):


    a,b = map(int,input().split())

    g = gcd(a,b)
    m = b//g
    b//=g
    l = []
    for i in range(2,int(sqrt(m)) + 1):

        if b%i==0:
            l.append(i)
            while(b%i==0):
                b//=i

    if b>1:
        l.append(b)
    ans = 0
    for i in l:
        m*=(i-1)
        m//=i
    print(m)

