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.
from math import sqrt
import sys


def putin():
    return map(int, sys.stdin.readline().split())


def sol():
    n = int(input())
    G = []
    for i in range(n):
        G.append(set())
    for i in range(n - 1):
        a, b = putin()
        a -= 1
        b -= 1
        G[a].add(b)
        G[b].add(a)
    newG = []
    for elem in G:
        newG.append(elem.copy())
    visited = [0] * n
    stack = [(0, 0)]
    to_del = []
    while len(stack) > 0:
        cur, parent = stack[-1]
        if visited[cur] == 0:
            visited[cur] = 1
            for elem in G[cur]:
                if elem != parent:
                    stack.append((elem, cur))
        else:
            stack.pop()
            visited[cur] = 2
            if parent != cur and len(newG[cur]) > 2 and len(newG[parent]) > 2:
                newG[cur].remove(parent)
                newG[parent].remove(cur)
                to_del.append((cur, parent))
    for i in range(n):
        while len(newG[i]) > 2:
            AA = newG[i].pop()
            newG[AA].remove(i)
            to_del.append((i, AA))
    hang = set()
    for i, elem in enumerate(newG):
        if len(elem) <= 1:
            hang.add(i)
    pairs = []
    while len(hang) > 0:
        A = hang.pop()
        AA = A
        while len(newG[A]) > 0:
            B = newG[A].pop()
            newG[B].remove(A)
            A = B
        if A in hang:
            hang.remove(A)
        pairs.append((AA, A, hang))
    print(len(to_del))
    for i in range(len(to_del)):
        print(to_del[i][0] + 1, to_del[i][1] + 1, pairs[i][1] + 1, pairs[i + 1][0] + 1)


for iter in range(int(sys.stdin.readline())):
    sol()
