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.

<image>

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

<image>

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:
"""
    Satwik_Tiwari ;) .
    12th Sept , 2020  - Saturday
"""

#===============================================================================================
#importing some useful libraries.
from __future__ import division, print_function

from fractions import Fraction
import sys
import os
from io import BytesIO, IOBase

from itertools import *
import bisect
from heapq import *
from math import *
from copy import *
from collections import deque
from collections import Counter as counter  # Counter(list)  return a dict with {key: count}
from itertools import combinations as comb # if a = [1,2,3] then print(list(comb(a,2))) -----> [(1, 2), (1, 3), (2, 3)]
from itertools import permutations as permutate
from bisect import bisect_left as bl

#If the element is already present in the list,
# the left most position where element has to be inserted is returned.

from bisect import bisect_right as br
from bisect import bisect

#If the element is already present in the list,
# the right most position where element has to be inserted is returned

#==============================================================================================
#fast I/O region
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")


def print(*args, **kwargs):
    """Prints the values to a stream, or to sys.stdout by default."""
    sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
    at_start = True
    for x in args:
        if not at_start:
            file.write(sep)
        file.write(str(x))
        at_start = False
    file.write(kwargs.pop("end", "\n"))
    if kwargs.pop("flush", False):
        file.flush()


if sys.version_info[0] < 3:
    sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
    sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)

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

#===============================================================================================
### START ITERATE RECURSION ###
from types import GeneratorType
def iterative(f, stack=[]):
  def wrapped_func(*args, **kwargs):
    if stack: return f(*args, **kwargs)
    to = f(*args, **kwargs)
    while True:
      if type(to) is GeneratorType:
        stack.append(to)
        to = next(to)
        continue
      stack.pop()
      if not stack: break
      to = stack[-1].send(to)
    return to
  return wrapped_func
#### END ITERATE RECURSION ####

#===============================================================================================
#some shortcuts

mod = 1000000007
def inp(): return sys.stdin.readline().rstrip("\r\n") #for fast input
def out(var): sys.stdout.write(str(var))  #for fast output, always take string
def lis(): return list(map(int, inp().split()))
def stringlis(): return list(map(str, inp().split()))
def sep(): return map(int, inp().split())
def strsep(): return map(str, inp().split())
# def graph(vertex): return [[] for i in range(0,vertex+1)]
def zerolist(n): return [0]*n
def nextline(): out("\n")  #as stdout.write always print sring.
def testcase(t):
    for pp in range(t):
        solve(pp)
def printlist(a) :
    for p in range(0,len(a)):
        out(str(a[p]) + ' ')
def google(p):
    print('Case #'+str(p)+': ',end='')
def lcm(a,b): return (a*b)//gcd(a,b)
def power(x, y, p) :
    res = 1     # Initialize result
    x = x % p  # Update x if it is more , than or equal to p
    if (x == 0) :
        return 0
    while (y > 0) :
        if ((y & 1) == 1) : # If y is odd, multiply, x with result
            res = (res * x) % p

        y = y >> 1      # y = y/2
        x = (x * x) % p
    return res
def ncr(n,r): return factorial(n)//(factorial(r)*factorial(max(n-r,1)))
def isPrime(n) :
    if (n <= 1) : return False
    if (n <= 3) : return True
    if (n % 2 == 0 or n % 3 == 0) : return False
    i = 5
    while(i * i <= n) :
        if (n % i == 0 or n % (i + 2) == 0) :
            return False
        i = i + 6
    return True

#===============================================================================================
# code here ;))
@iterative
def dfs(v,visited,st,sub,par):
    visited[st] = 1
    for i in v[st]:
        if(visited[i] == 0):
            ok = yield dfs(v,visited,i,sub,par)
            sub[st]+=sub[i]
            par[i] = st

    yield True

@iterative
def dfslol(v,visited,st,a,edeg):
    # if(len(edeg)!=0):
    #     return
    # print(st)
    visited[st] = 1
    for i in v[st]:
        if(visited[i] == 0 and i!=a):
            ok = yield dfslol(v,visited,i,a,edeg)
            if(len(v[i]) == 1 and len(edeg)==0):
                edeg.append(st)
                edeg.append(i)
            # print(st,sub)
    yield True

def solve(case):
    n = int(inp())
    g = [[] for i in range(n)]
    help =[]
    for i in range(n-1):
        x,y = sep()
        x-=1
        y-=1
        if(len(help) == 0):
            help.append(x)
            help.append(y)
        g[x].append(y)
        g[y].append(x)

    sub = [1]*n
    vis = [0]*n
    par = [-1]*n
    # print(g)
    dfs(g,vis,0,sub,par)
    # print(sub)
    if(n%2==1):
        print(help[0]+1,help[1]+1)
        print(help[0]+1,help[1]+1)
        return

    centroids = []
    maxmin = 10**10
    for i in range(n):
        curr = 0
        for j in g[i]:
            if(j == par[i]):
                curr = max(curr,n - sub[i])
            else:
                curr = max(curr,sub[j])
        maxmin = min(maxmin,curr)
    # print(maxmin)
    for i in range(n):
        curr = 0
        for j in g[i]:
            if(j == par[i]):
                curr = max(curr,n - sub[i])
            else:
                curr = max(curr,sub[j])
        if(curr == maxmin):
            centroids.append(i)

    if(len(centroids) == 1):
        print(help[0]+1,help[1]+1)
        print(help[0]+1,help[1]+1)
        return

    vis = [0]*n
    edge = []
    dfslol(g,vis,centroids[1],centroids[0],edge)
    # print(edge,centroids)
    print(edge[0]+1,edge[1]+1)
    print(centroids[0]+1,edge[1]+1)

# 6
# 0 1
# 0 3
# 1 2
# 2 3
# 3 4
# 4 5





# testcase(1)
testcase(int(inp()))