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.
def order(n,G):
    ans = []
    parent = [-1]*n

    def dfs(x,par):
        parent[x] = par
        for y in G[x]:
            if y == par:
                continue
            dfs(y,x)
        ans.append(x)
    
    dfs(1,-1)
    return ans,parent

def solve(n,G,topo,parent):
    to_remove = []

    for x in topo:
        p = parent[x]
        num_child = len(G[x]) - (1 if p in G[x] else 0)

        if num_child <= 1:
            pass
        else:
            if p != -1:
                to_remove.append((x,p))
                G[x].remove(p)
                G[p].remove(x)

            to_select = 0
            tmp = []
            
            for y in G[x]:
                if y == p:
                    continue
                if to_select < 2:
                    to_select += 1
                else:
                    to_remove.append((y,x))
                    tmp.append(y)

            for y in tmp:
                G[x].remove(y)
                G[y].remove(x)

    visited = [False]*n
    def find_bamboo(x):
        visited[x] = True

        if not G[x]:
            return x,x

        par = x
        while len(G[x]) > 1:
            for y in G[x]:
                if y != par:
                    par = x
                    x = y
                    break
        par = x
        cur = list(G[x])[0]
        visited[x] = True
        visited[cur] = True
        while len(G[cur]) > 1:
            for y in G[cur]:
                if y != par:
                    par = cur
                    cur = y
                    visited[y] = True
                    break
        return x,cur

    bamboos = []

    for x in range(n):
        if not visited[x]:
            u,v = find_bamboo(x)
            bamboos.append((u,v))
    
    k = len(to_remove)
    print(k)
    for i in range(k):
        a,b = bamboos[i][1], bamboos[i+1][0]
        u,v = to_remove[i]

        print(u+1,v+1,a+1,b+1)

    return




for nt in range(int(input())):
    n = int(input())
    G = [set() for _ in range(n)]
    for _ in range(n-1):
        a,b = map(int,input().split())
        G[a-1].add(b-1)
       

        G[b-1].add(a-1)

    topo,parent = order(n,G)
    solve(n,G,topo,parent)
