You are given an unweighted tree with n vertices. Recall that a tree is a connected undirected graph without cycles.

Your task is to choose three distinct vertices a, b, c on this tree such that the number of edges which belong to at least one of the simple paths between a and b, b and c, or a and c is the maximum possible. See the notes section for a better understanding.

The simple path is the path that visits each vertex at most once.

Input

The first line contains one integer number n (3 ≤ n ≤ 2 ⋅ 10^5) — the number of vertices in the tree. 

Next n - 1 lines describe the edges of the tree in form a_i, b_i (1 ≤ a_i, b_i ≤ n, a_i ≠ b_i). It is guaranteed that given graph is a tree.

Output

In the first line print one integer res — the maximum number of edges which belong to at least one of the simple paths between a and b, b and c, or a and c.

In the second line print three integers a, b, c such that 1 ≤ a, b, c ≤ n and a ≠, b ≠ c, a ≠ c.

If there are several answers, you can print any.

Example

Input


8
1 2
2 3
3 4
4 5
4 6
3 7
3 8


Output


5
1 8 6

Note

The picture corresponding to the first example (and another one correct answer):

<image>

If you choose vertices 1, 5, 6 then the path between 1 and 5 consists of edges (1, 2), (2, 3), (3, 4), (4, 5), the path between 1 and 6 consists of edges (1, 2), (2, 3), (3, 4), (4, 6) and the path between 5 and 6 consists of edges (4, 5), (4, 6). The union of these paths is (1, 2), (2, 3), (3, 4), (4, 5), (4, 6) so the answer is 5. It can be shown that there is no better answer.
#3 times dfs
#first time to find the longest leaf from the random root
#second time to find the longest path in the tree
#third time to determine the answer
#mind the special case
from types import GeneratorType
 
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

import sys
from collections import deque
import random
input = sys.stdin.readline
n = int(input().strip())
tree = [[] for i in range(n + 1)]
for i in range(n - 1):
    a,b = map(int,input().strip().split())
    tree[a].append(b)
    tree[b].append(a)

#first bfs
q = deque()
root = random.randrange(1,n + 1);max_dep = 0;max_node = root
for node in tree[root]:
    q.append((node,root,1))

while len(q) > 0:
    node,pd,dep = q.popleft()
    if len(tree[node]) <= 1:
        if dep > max_dep:
            max_node = node
            max_dep = dep

    for child in tree[node]:
        if child == pd:
            continue
        q.append((child,node,dep + 1))


#second bfs  
pa = [0 for i in range(n + 1)];pa[root] = root
root = max_node; se = 0; max_dep = 0; first = max_node
for node in tree[root]:
    q.append((node,root,1))
    
while len(q) > 0:
    node,pd,dep = q.popleft()
    pa[node] = pd
    if len(tree[node]) <= 1:
        if dep > max_dep:
            se = node
            max_dep = dep

    for child in tree[node]:
        if child == pd:
            continue
        q.append((child,node,dep + 1))

#another dfs
ans = max_dep
max_dep = [0]
max_node = [se]
@bootstrap
def dfs(node,pd,depth,desc):
    if len(tree[node]) <= 1:
        if depth > max_dep[0]:
            max_dep[0] = depth
            max_node[0] = node
        yield max_dep

    for child in tree[node]:
        if child == pd or child == desc:
            continue
        yield dfs(child,node,depth + 1,desc)

    yield max_dep

to_dfs = pa[se];desc = se
while pa[to_dfs] != to_dfs:
    #print(to_dfs,pa[to_dfs])
    dfs(to_dfs,pa[to_dfs],0,desc)
    desc = to_dfs
    to_dfs = pa[to_dfs]
    
third = max_node[0]
print(ans + max_dep[0])
if third == se:
    print(first,se,pa[se])
else:
    print(first,se,third)
