In 1976 the “Four Color Map Theorem” was proven with the assistance of a computer. This theorem
states that every map can be colored using only four colors, in such a way that no region is colored
using the same color as a neighbor region.

Here you are asked to solve a simpler similar problem. You have to decide whether a given arbitrary
connected graph can be bicolored. That is, if one can assign colors (from a palette of two) to the nodes
in such a way that no two adjacent nodes have the same color. To simplify the problem you can assume:

• no node will have an edge to itself.

• the graph is nondirected. That is, if node a is said to be connected to node b, then you must
assume that b is connected to a.

• the graph will be strongly connected. That is, there will be at least one path from any node to
any other node.

Input

First line of input contains the total number of test cases T.
Each test case starts with a line containing the number n (1 < n < 200) of different nodes. The second line contains the number of edges l. After this, l lines will follow, each containing two numbers that specify an edge between the two nodes that they represent. A node in the graph will be labeled using a number a (0 ≤ a < n).

Output

You have to decide whether the input graph can be bicolored or not, and print it as shown below.
If it can be bicolored then print "BICOLORABLE." else "NOT BICOLORABLE." (without quotes).

Constraints

1 ≤ t ≤ 10
1 ≤ n < 200 
1 ≤ n ≤ 10^4

SAMPLE INPUT
1
3
3
0 1
1 2
2 0

SAMPLE OUTPUT
NOT BICOLORABLE.
'''
# Read input from stdin and provide input before running code

name = raw_input('What is your name?\n')
print 'Hi, %s.' % name
'''

def bfs(arr,start,colors,curr_col):
	
	ans=True
	next_i=[]
	
	for a in arr[start] :
		
		if(colors[a] == -1):
			colors[a]=curr_col
			next_i += [a]
			
			
		elif(colors[a]!= curr_col):
			ans=False
	
	
	for a in next_i:
		ans= ans and bfs(arr,a,colors,curr_col^1)
	
	return ans
	

T=int(raw_input())

for t in range(T):
	
	n=int(raw_input())
	
	m=int(raw_input())
	
	arr=[]
	for i in range(n):
		arr=arr+ [[]]
		
	
	
	for i in range(m):
	
		s=raw_input().split(" ")
		a=int(s[0])
		b=int(s[1])
		
		arr[a] += [b]
		arr[b] += [a]
		
	#print(arr)	
	
	
	colors=[-1]*n
	
	ans=(bfs(arr,0,colors,0))
	
	if(ans):
		print("BICOLORABLE.")
	else:
		print("NOT BICOLORABLE.")