Problem: Fishing Prince loves trees, and he especially loves trees with only one centroid. The tree is a connected graph without cycles.

A vertex is a centroid of a tree only when you cut this vertex (remove it and remove all edges from this vertex), the size of the largest connected component of the remaining graph is the smallest possible.

For example, the centroid of the following tree is 2, because when you cut it, the size of the largest connected component of the remaining graph is 2 and it can't be smaller.

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However, in some trees, there might be more than one centroid, for example:

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Both vertex 1 and vertex 2 are centroids because the size of the largest connected component is 3 after cutting each of them.

Now Fishing Prince has a tree. He should cut one edge of the tree (it means to remove the edge). After that, he should add one edge. The resulting graph after these two operations should be a tree. He can add the edge that he cut.

He wants the centroid of the resulting tree to be unique. Help him and find any possible way to make the operations. It can be proved, that at least one such way always exists.

Input

The input consists of multiple test cases. The first line contains an integer t (1≤ t≤ 10^4) — the number of test cases. The description of the test cases follows.

The first line of each test case contains an integer n (3≤ n≤ 10^5) — the number of vertices.

Each of the next n-1 lines contains two integers x, y (1≤ x,y≤ n). It means, that there exists an edge connecting vertices x and y.

It's guaranteed that the given graph is a tree.

It's guaranteed that the sum of n for all test cases does not exceed 10^5.

Output

For each test case, print two lines.

In the first line print two integers x_1, y_1 (1 ≤ x_1, y_1 ≤ n), which means you cut the edge between vertices x_1 and y_1. There should exist edge connecting vertices x_1 and y_1.

In the second line print two integers x_2, y_2 (1 ≤ x_2, y_2 ≤ n), which means you add the edge between vertices x_2 and y_2.

The graph after these two operations should be a tree.

If there are multiple solutions you can print any.

Example

Input


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


Output


1 2
1 2
1 3
2 3

Note

Note that you can add the same edge that you cut.

In the first test case, after cutting and adding the same edge, the vertex 2 is still the only centroid.

In the second test case, the vertex 2 becomes the only centroid after cutting the edge between vertices 1 and 3 and adding the edge between vertices 2 and 3.

Solution (in PYTHON3):

Solution:
import os
import sys
from io import BytesIO, IOBase

_str = str
BUFSIZE = 8192


def str(x=b''):
    return x if type(x) is bytes else _str(x).encode()


class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = 'x' in file.mode or 'r' not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b'\n') + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)


class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode('ascii'))
        self.read = lambda: self.buffer.read().decode('ascii')
        self.readline = lambda: self.buffer.readline().decode('ascii')


sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)


def inp():
    return sys.stdin.readline().rstrip('\n')


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


def itg():
    return int(sys.stdin.readline())

# ############################## import
from copy import deepcopy


def to_tree(graph, root=0):
    """
    graph: undirected graph (adjacency list)
    :return directed graph that parent -> children
    """
    graph[:] = map(set, graph)

    stack = [[root]]
    while stack:
        if not stack[-1]:
            del stack[-1]
            continue
        vertex = stack[-1].pop()
        for e in graph[vertex]:
            graph[e].remove(vertex)
        stack.append(list(graph[vertex]))

    graph[:] = map(list, graph)


def to_graph(tree, to_list=False):
    """ :return undirected graph (adjacency list) """
    tree[:] = map(set, tree)
    for node1, node_set in enumerate(deepcopy(tree)):
        for node2 in node_set:
            tree[node2].add(node1)
    if to_list:
        tree[:] = map(list, tree)


def tree_bfs(tree, start=0, flat=False) -> list:
    stack = [start]
    result = []
    while stack:
        new_stack = []
        if flat:
            result.extend(stack)
        else:
            result.append(stack)
        for node in stack:
            new_stack.extend(tree[node])
        stack = new_stack
    return result


def tree_postorder(tree, root=0, flat=False):
    return tree_bfs(tree, root, flat=flat)[::-1]


def tree_edge_descendant(tree, edge_time_count=False, root=0):
    """
    :return: a list for the number of vertices under each edge

    if edge_time_count:
    :return a list for the sum of times that walking on each edge
    the walk mean the path of any pair (u, v) and (u < v)
    """
    vertices = len(tree)
    graph = deepcopy(tree)
    to_graph(graph)
    results = [[] for _ in range(vertices)]
    for leaves in tree_postorder(tree)[:-1]:
        for leaf in leaves:
            graph[leaf] = list(graph[leaf])
            graph[graph[leaf][0]].remove(leaf)
        for leaf in leaves:
            children = sum(results[leaf]) + 1
            results[graph[leaf][0]].append(children)

    if edge_time_count:
        for result in results:
            result[:] = map(lambda x: x * (vertices - x), result)
    return results


def tree_split_count(tree):
    """
    return a list for each node, which contains
    the size of subtree if take off the node
    """
    vertices = len(tree)
    edge_descendant = tree_edge_descendant(tree)
    for descendant in edge_descendant:
        other = vertices - sum(descendant) - 1
        if other:
            descendant.append(other)
    return edge_descendant


def tree_centroid(tree) -> list:
    vertices = len(tree)
    half = vertices >> 1
    split_count = tree_split_count(tree)
    centroid = []
    for node, count in enumerate(split_count):
        if max(count) <= half:
            centroid.append(node)
            if vertices & 1 or len(centroid) == 2:
                break
    return centroid


# ############################## main
for __ in range(itg()):
    n = itg()
    graph = [[] for _ in range(n)]
    if n & 1:
        for _ in range(n - 2):
            inp()
        ans = inp()
        print(ans)
        print(ans)
        continue

    for _ in range(n - 1):
        u, v = mpint()
        u -= 1
        v -= 1
        graph[u].append(v)
        graph[v].append(u)

    tree = deepcopy(graph)
    to_tree(tree)
    centroids = tree_centroid(tree)
    if len(centroids) == 1:
        c1 = centroids[0]
        print(c1 + 1, graph[c1][0] + 1)
        print(c1 + 1, graph[c1][0] + 1)
        continue
    c1, c2 = centroids
    if c2 in tree[c1]:
        c1, c2 = c2, c1
    parent = c1
    leaf = tree[c1][0]
    while tree[leaf]:  # leaf is not leaf
        parent = leaf
        leaf = tree[leaf][0]
    print(parent + 1, leaf + 1)
    print(c2 + 1, leaf + 1)

# Please check!