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 sys
import math
import heapq
import bisect
from collections import Counter
from collections import defaultdict
from io import BytesIO, IOBase
from types import GeneratorType
import string


class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        import os
        self.os = os
        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
        self.BUFSIZE = 8192

    def read(self):
        while True:
            b = self.os.read(self._fd, max(self.os.fstat(self._fd).st_size, self.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 = self.os.read(self._fd, max(self.os.fstat(self._fd).st_size, self.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:
            self.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")


def get_int():
    return int(input())


def get_ints():
    return list(map(int, input().split(' ')))


def get_int_grid(n):
    return [get_ints() for _ in range(n)]


def get_str():
    return input().strip()


def get_strs():
    return get_str().split(' ')


def flat_list(arr):
    return [item for subarr in arr for item in subarr]


def yes_no(b):
    if b:
        return "YES"
    else:
        return "NO"


def binary_search(good, left, right, delta=1, right_true=False):
    """
    Performs binary search
    ----------
    Parameters
    ----------
    :param good: Function used to perform the binary search
    :param left: Starting value of left limit
    :param right: Starting value of the right limit
    :param delta: Margin of error, defaults value of 1 for integer binary search
    :param right_true: Boolean, for whether the right limit is the true invariant
    :return: Returns the most extremal value interval [left, right] which is good function evaluates to True,
            alternatively returns False if no such value found
    """

    limits = [left, right]
    while limits[1] - limits[0] > delta:
        if delta == 1:
            mid = sum(limits) // 2
        else:
            mid = sum(limits) / 2
        if good(mid):
            limits[int(right_true)] = mid
        else:
            limits[int(~right_true)] = mid
    if good(limits[int(right_true)]):
        return limits[int(right_true)]
    else:
        return False


def prefix_sums(a):
    p = [0]
    for x in a:
        p.append(p[-1] + x)
    return p


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


def solve_a():
    n = get_int()
    a = get_ints()
    ans = 0
    for i in range(n + 1):
        if a.count(i) < 2:
            ans += i
            break
    for i in range(n + 1):
        if a.count(i) < 1:
            ans += i
            break
    return ans


def solve_b():
    n = get_int()
    a = get_ints()
    a.sort()
    x = []
    x.append(a[0] * a[1] * a[2] * a[3] * a[-1])
    x.append(a[0] * a[1] * a[-3] * a[-2] * a[-1])
    x.append(a[-5] * a[-4] * a[-3] * a[-2] * a[-1])
    return max(x)


def centroid(T):
    n = len(T)
    sz = [0] * n
    centroid = []
    vis = [False] * n

    @bootstrap
    def dfs(u):
        vis[u] = True
        sz[u] = 1
        is_centroid = True
        for v in T[u]:
            if vis[v]:
                continue
            yield dfs(v)
            sz[u] += sz[v]
            if sz[v] > n / 2:
                is_centroid = False
        if n - sz[u] > n / 2:
            is_centroid = False
        if is_centroid:
            centroid.append(u)
        yield
    dfs(0)
    return centroid


def get_tree():
    n = get_int()
    T = [[] for _ in range(n)]
    for _ in range(n - 1):
        a, b = get_ints()
        a -= 1
        b -= 1
        T[a].append(b)
        T[b].append(a)
    return T


def solve_c():
    T = get_tree()
    c = centroid(T)
    if len(c) == 1:
        edge = (c[0] + 1, T[c[0]].pop() + 1)
        return edge, edge
    elif len(c) == 2:
        vis = [False] * len(T)
        vis[c[0]] = True
        leafs = []

        @bootstrap
        def dfs2(u):
            if len(T[u]) == 1:
                leafs.append(u)
                yield
            vis[u] = True
            for v in T[u]:
                if vis[v]:
                    continue
                else:
                    yield dfs2(v)
            yield

        dfs2(c[1])
        leaf = leafs[0]
        edge1 = (leaf + 1, T[leaf].pop() + 1)
        edge2 = (leaf + 1, c[0] + 1)
        return edge1, edge2


t = get_int()
for _ in range(t):
    ans = solve_c()
    for edge in ans:
        print(*edge)