fredos-crush-2
Source: hackerearth
Difficulty: MEDIUM

Fredo has crush on a girl in his class. He went to ask her out for a date. She, being a coder, gave her a simple graph with N vertices and M edges and asked him to solve a problem on the graph in order to go on a date with her. 
The problem says:
Let there be some functions defined below:
For each vertex in the graph:
S = max (sum of all vertices lying in the maximum length simple path ending at that vertex)
That is, if there are two simple paths with length 3 ending at some vertex and there can be no simple path with length greater than 3, and P be the sum of vertices of first path and Q of second. So, S = max(P,Q).
Let A = minimum value of S[i] (1 ≤ i ≤ n)
B = maximum value of S[i] (1 ≤ i ≤ n)
You are given an equation Ax = By 
Find the minimum value of x and y satisfying the equation.
Note: If there are no simple paths ending at some vertex, S for that vertex will be number of that vertex.
A simple path is a path in a graph which does not have repeating vertices.

Input Format:

First line contains an integer T denoting the number of test cases.
First line of each test case consists of  two space separated integers denoting N and M and the following M lines consist of two space separated integers X and Y denoting there is an edge between vertices X and Y.  

Output Format:

For each test case, print the values of x and y (space separated) in a separate line.

Constraints:
 1 ≤ T ≤ 10 
 1 ≤ N ≤ 15 
 0 ≤ M < N*(N-1)/2  
 1 ≤ X, Y ≤ N 

SAMPLE INPUT
2
3 0
3 2
1 2
2 3

SAMPLE OUTPUT
3 1
6 5

Explanation

In the first case, there are no simple paths ending at any vertex, so S={1,2,3}, we need to solve for 1 * x=3 * y.
In the second case, 
paths ending at 1 are: 3-2-1 and 2-1. The maximum length path is 3-2-1 having sum=3+2+1=6
paths ending at 2 are: 1-2 and 3-2. Both have the same length, s[2]=max(1+2,3+2)=5
and similarly for 3, s[3]=6
so we need to solve for 5 * x=6 * y.