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
from fractions import gcd
import math
for _ in xrange(int(raw_input())):
	a,b,d=map(int,raw_input().split());	A=[];
	q=gcd(a,b);	s=int(math.sqrt(q));
	for i in xrange(1,s+1):
		if q%i==0:
			A.append(i);	A.append(q/i);
	if s*s==q:
		A.pop()
	A.sort(reverse=True);
	if len(A)<d:
		print "No crymeth today"
	else:
		print A[d-1]