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:
# ------------------- fast io --------------------
import os
import sys
from io import BytesIO, IOBase

BUFSIZE = 8192


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)
input = lambda: sys.stdin.readline().rstrip("\r\n")

# ------------------- fast io --------------------
from math import gcd, ceil

def pre(s):
    n = len(s)
    pi = [0] * n
    for i in range(1, n):
        j = pi[i - 1]
        while j and s[i] != s[j]:
            j = pi[j - 1]
        if s[i] == s[j]:
            j += 1
        pi[i] = j
    return pi


def prod(a):
    ans = 1
    for each in a:
        ans = (ans * each)
    return ans

def lcm(a, b): return a * b // gcd(a, b)


def binary(x, length=16):
    y = bin(x)[2:]
    return y if len(y) >= length else "0" * (length - len(y)) + y


def centroid_decomposition(graph):
    """
    Given a tree, this function is a generator that
    1. Roots the tree at its centroid (modifying graph)
    2. Yields centroid
    3. Removes centroid from the graph
    4. Recurses on the forest left after the removal

    This generator makes working with centroid decomposition easy. It yields the n
    centroids involved in the decomposition. It also keeps the graph rooted at the yielded
    centroid by modifying the input variable graph. In total this takes O(n log n) time.
    Input:
      graph: list of lists where graph[u] is a list containing all neighbours of node u

    Example:
      >>> graph = [[1], [0,2], [1]]
      >>> for centroid in centroid_decomposition(graph):
      >>>   bfs = [centroid]
      >>>   for node in bfs:
      >>>     bfs += graph[node] # Valid since graph is rooted at the centroid
      >>>   print('BFS from centroid:', bfs)
      BFS from centroid: [1, 0, 2]
      BFS from centroid: [0]
      BFS from centroid: [2]
    """
    n = len(graph)

    bfs = [n - 1]
    for node in bfs:
        bfs += graph[node]
        for nei in graph[node]:
            graph[nei].remove(node)

    size = [0] * n
    for node in reversed(bfs):
        size[node] = 1 + sum(size[child] for child in graph[node])

    def reroot_centroid(root):
        N = size[root]
        while True:
            for child in graph[root]:
                if size[child] > N // 2:
                    size[root] = N - size[child]
                    graph[root].remove(child)
                    graph[child].append(root)
                    root = child
                    break
            else:
                return root

    bfs = [n - 1]
    for node in bfs:
        centroid = reroot_centroid(node)
        bfs += graph[centroid]
        return centroid

def dfs(graph, start=0):
    n = len(graph)

    dp = [0] * n
    visited, finished = [False] * n, [False] * n

    stack = [start]
    while stack:
        start = stack[-1]

        # push unvisited children into stack
        if not visited[start]:
            visited[start] = True
            for child in graph[start]:
                if not visited[child]:
                    stack.append(child)

        else:
            stack.pop()

            # base case
            dp[start] += 1

            # update with finished children
            for child in graph[start]:
                if finished[child]:
                    dp[start] += dp[child]

            finished[start] = True

    return visited, dp

for _ in range(int(input()) if True else 1):
    n = int(input())
    #n, k = map(int, input().split())
    #a, b = map(int, input().split())
    #c, d = map(int, input().split())
    #a = list(map(int, input().split()))
    #b = list(map(int, input().split()))
    #s = input()
    graph = [[] for i in range(n+1)]
    for i in range(n-1):
        x, y = map(int, input().split())
        graph[x] += [y]
        graph[y] += [x]
    graph2 = [list(graphh) for graphh in graph]
    centroid1 = centroid_decomposition(graph)
    graph = [list(k) for k in graph2]
    visited, dp = dfs(graph, centroid1)
    x = dp[centroid1]
    dp[centroid1] = 0
    ma = max(dp)
    if 2*ma == x:
        centroid2 = dp.index(ma)
        for i in graph[centroid1]:
            if i != centroid2:
                print(i, centroid1)
                print(i, centroid2)
                break
    else:
        print(1, graph[1][0])
        print(1, graph[1][0])