Nastia has an unweighted tree with n vertices and wants to play with it!

The girl will perform the following operation with her tree, as long as she needs:

  1. Remove any existing edge. 
  2. Add an edge between any pair of vertices. 



What is the minimum number of operations Nastia needs to get a bamboo from a tree? A bamboo is a tree in which no node has a degree greater than 2.

Input

The first line contains a single integer t (1 ≤ t ≤ 10 000) — the number of test cases.

The first line of each test case contains a single integer n (2 ≤ n ≤ 10^5) — the number of vertices in the tree.

Next n - 1 lines of each test cases describe the edges of the tree in form a_i, b_i (1 ≤ a_i, b_i ≤ n, a_i ≠ b_i).

It's guaranteed the given graph is a tree and the sum of n in one test doesn't exceed 2 ⋅ 10^5.

Output

For each test case in the first line print a single integer k — the minimum number of operations required to obtain a bamboo from the initial tree.

In the next k lines print 4 integers x_1, y_1, x_2, y_2 (1 ≤ x_1, y_1, x_2, y_{2} ≤ n, x_1 ≠ y_1, x_2 ≠ y_2) — this way you remove the edge (x_1, y_1) and add an undirected edge (x_2, y_2).

Note that the edge (x_1, y_1) must be present in the graph at the moment of removing.

Example

Input


2
7
1 2
1 3
2 4
2 5
3 6
3 7
4
1 2
1 3
3 4


Output


2
2 5 6 7
3 6 4 5
0

Note

Note the graph can be unconnected after a certain operation.

Consider the first test case of the example: 

<image> The red edges are removed, and the green ones are added.
'''Author- Akshit Monga'''
# Don't know the proper proof
from sys import stdin, stdout
input = stdin.readline

from types import GeneratorType
def bootstrap(f, stack=[]):
    def wrappedfunc(*args, **kwargs):
        if stack:
            return f(*args, **kwargs)
        else:
            to = f(*args, **kwargs)
            while True:
                if type(to) is GeneratorType:
                    stack.append(to)
                    to = next(to)
                else:
                    stack.pop()
                    if not stack:
                        break
                    to = stack[-1].send(to)
            return to
    return wrappedfunc

@bootstrap
def detach(start,parent):
    global removed
    childs=list(graph[start])
    for i in childs:
        if i==parent:
            continue
        yield detach(i,start)
    childs=graph[start].copy()
    if parent:
        childs.remove(parent)
    childs=list(childs)
    if len(childs)>=2:
        if parent:
            removed.append((start,parent))
            graph[start].remove(parent)
            graph[parent].remove(start)
        for i in childs[2:]:
            removed.append((start,i))
            graph[start].remove(i)
            graph[i].remove(start)
    yield

@bootstrap
def dfs(start):
    global check
    vis.add(start)
    count=0
    for i in graph[start]:
        count+=1
        if i not in vis:
            yield dfs(i)
    if count==0:
        leaves.append(start)
        leaves.append(start)
    elif count==1:
        leaves.append(start)
    yield


t = int(input())
for _ in range(t):
    n=int(input())
    graph=[set() for i in range(n+1)]
    for i in range(n-1):
        a,b=map(int,input().split())
        graph[a].add(b)
        graph[b].add(a)
    removed=[]
    added=[]
    leaves=[]
    detach(1,0)
    vis=set()
    for i in range(1,n+1):
        if i not in vis:
            dfs(i)
    for i in range(1,len(leaves)-1,2):
        added.append((leaves[i],leaves[i+1]))
    x=len(removed)
    print(x)
    for i in range(x):
        stdout.write(str(removed[i][0])+" "+str(removed[i][1])+" "+str(added[i][0])+" "+str(added[i][1])+'\n')