Problem: Walter and Jesse's friend Mike had helped them in making Crymeth and hence, they wanted to give him a share.
For deciding the share, they both decided to choose one number each, X and Y and found out that K^th Highest Common Factor of their two numbers is a good amount of Crymeth that can be given to Mike .
Walter and Jesse did this activity for D days in total. For each day, find out what quantity of Crymeth they decided to give to Mike.

Input:

First line contains a natural number D - the total number of days.
D lines follow. Each line contains three natural numbers - X, Y and K.

Output:

For each day, print the quantity given to Mike on a new line.
If the K^th HCF of X and Y does not exist, simply print "No crymeth today" (Without the quotes)

Constraints:
1 ≤ D ≤ 10^3
1 ≤ X, Y, K ≤ 10^10

SAMPLE INPUT
3
8 16 2
8 16 4
8 16 5

SAMPLE OUTPUT
4
1
No crymeth today

Explanation

1st Highest Common Factor of 8 and 16 = 8
2nd Highest Common Factor of 8 and 16 = 4
3rd Highest Common Factor of 8 and 16 = 2
4th Highest Common Factor of 8 and 16 = 1

Solution (in PYTHON):

Solution:
def hcf(x, y):

   while(y):
       x, y = y, x % y
   return x
def factors(n):    
	return list(set(reduce(list.__add__, ([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0))))
inp1 = raw_input()
for i in range(int(inp1)):
	inp = map(int,raw_input().split(' '))
	if inp[0] > inp[1]:
		inp[0],inp[1] = inp[1],inp[0]
	hcfs = hcf(inp[0],inp[1])
	lis = factors(hcfs)
	sortlist = sorted(lis, reverse=True)
	if inp[2]-1 < len(sortlist):
		print sortlist[inp[2]-1]
	else:
		print "No crymeth today"