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 popcnt(n):
	return bin(n).count("1") % 2

def calc(gen, limit):
	divisors = []
	i = 2
	gen_copy = gen
	while gen > 1 and i*i <= gen_copy:
		if gen%i == 0:
			divisors.append(i)
			while gen%i == 0:
				gen //= i
		i += 1
	if gen != 1:
		divisors.append(gen)
	#print(divisors)

	l = len(divisors)
	res = 0
	for bit in range(1 << l):
		prod = 1
		for j, div in enumerate(divisors):
			if (bit>>j) & 1:
				prod *= div
		#print(bit, prod)
		if popcnt(bit):
			res -= (limit-1) // prod
		else:
			res += (limit-1) // prod
	return res



T = int(input())
for _ in range(T):
	a, m = map(int, input().split())
	g = gcd(a, m)
	k, l = a//g, m//g
	print(calc(l, l+k) - calc(l, k))