Kefaa has developed a novel decomposition of a tree. He claims that this decomposition solves many difficult problems related to trees. However, he doesn't know how to find it quickly, so he asks you to help him.
You are given a tree with $N$ vertices numbered $1$ through $N$. Let's denote an edge between vertices $u$ and $v$ by $(u, v)$. The triple-tree decomposition is a partition of edges of the tree into unordered triples of edges $(a, b), (a, c), (a, d)$ such that $a \neq b \neq c \neq d$. Each edge must belong to exactly one triple.
Help Kefaa with this problem — find a triple-tree decomposition of the given tree or determine that no such decomposition exists.
-----Input-----
- The first line of the input contains a single integer $T$ denoting the number of test cases. The description of $T$ test cases follows.
- The first line of each test case contains a single integer $N$.
- Each of the following $N-1$ lines contains two space-separated integers $u$ and $v$ describing an edge between vertices $u$ and $v$ of the tree.
-----Output-----
- For each test case, print a line containing the string "YES" if a triple-tree decomposition of the given tree exists or "NO" otherwise.
- If it exists, print $\frac{N-1}{3}$ more lines describing a decomposition.
- Each of these lines should contain four space-separated integers $a$, $b$, $c$ and $d$ describing a triple of edges $(a, b), (a, c), (a, d)$.
If more than one triple-tree decomposition exists, you may output any one.
-----Constraints-----
- $1 \le T \le 100$
- $2 \le N \le 2 \cdot 10^5$
- $1 \le u, v \le N$
- the sum of $N$ over all test cases does not exceed $2 \cdot 10^5$
-----Subtasks-----
Subtask #1 (20 points): $2 \le N \le 10$
Subtask #2 (30 points):$2 \le N \le 5000$ and the sum of $N$ overall testcases doesn't exceed $5000$
Subtask #3 (50 points): original constraints
-----Example Input-----
2
4
1 2
1 3
1 4
7
1 2
2 3
1 4
4 5
1 6
6 7
-----Example Output-----
YES
1 2 3 4
NO

test=int(input())
for t in range(test):
 n= int(input())

 adj=[[] for i in range(n+1)]

 for _ in range(n-1):
  a,b=list(map(int,input().split()))
  adj[a].append(b)
  adj[b].append(a)
 

 #print(adj)
 root=1
 q,s=[root],set([root])

 for x in q:
  adj[x]= [p for p in adj[x] if p not in s]
  q.extend(adj[x])
  s.update(adj[x])

 #print(adj)
 ans=True
 for i in range(n+1):
  if(len(adj[i]) %3!=0):
   ans=False
 if(ans):
  print("YES")
  for i in range(n+1):
   while(len(adj[i])):
    print(i,adj[i][0],adj[i][1],adj[i][2])
    adj[i].pop(0)
    adj[i].pop(0)
    adj[i].pop(0)
 else:
  print("NO")


test=int(input())
for t in range(test):
 n= int(input())

 adj=[[] for i in range(n+1)]

 for _ in range(n-1):
  a,b=list(map(int,input().split()))
  adj[a].append(b)
  adj[b].append(a)
 

 #print(adj)
 root=1
 q,s=[root],set([root])

 for x in q:
  adj[x]= [p for p in adj[x] if p not in s]
  q.extend(adj[x])
  s.update(adj[x])

 #print(adj)
 ans=True
 for i in range(n+1):
  if(len(adj[i]) %3!=0):
   ans=False
 if(ans):
  print("YES")
  for i in range(n+1):
   while(len(adj[i])):
    print(i,adj[i][0],adj[i][1],adj[i][2])
    adj[i].pop(0)
    adj[i].pop(0)
    adj[i].pop(0)
 else:
  print("NO")
