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.
class HLD:
    def __init__(self, g):
        self.g = g
        self.n = len(g)
        self.parent = [-1]*self.n
        self.size = [1]*self.n
        self.head = [0]*self.n
        self.preorder = [0]*self.n
        self.k = 0
        self.depth = [0]*self.n

        for v in range(self.n):
            if self.parent[v] == -1:
                self.dfs_pre(v)
                self.dfs_hld(v)

    def dfs_pre(self, v):
        g = self.g
        stack = [v]
        order = [v]
        while stack:
            v = stack.pop()
            for u in g[v]:
                if self.parent[v] == u:
                    continue
                self.parent[u] = v
                self.depth[u] = self.depth[v]+1
                stack.append(u)
                order.append(u)

        # 隣接リストの左端: heavyな頂点への辺
        # 隣接リストの右端: 親への辺
        while order:
            v = order.pop()
            child_v = g[v]
            if len(child_v) and child_v[0] == self.parent[v]:
                child_v[0], child_v[-1] = child_v[-1], child_v[0]
            for i, u in enumerate(child_v):
                if u == self.parent[v]:
                    continue
                self.size[v] += self.size[u]
                if self.size[u] > self.size[child_v[0]]:
                    child_v[i], child_v[0] = child_v[0], child_v[i]

    def dfs_hld(self, v):
        stack = [v]
        while stack:
            v = stack.pop()
            self.preorder[v] = self.k
            self.k += 1
            top = self.g[v][0]
            # 隣接リストを逆順に見ていく（親 > lightな頂点への辺 > heavyな頂点 (top)）
            # 連結成分が連続するようにならべる
            for u in reversed(self.g[v]):
                if u == self.parent[v]:
                    continue
                if u == top:
                    self.head[u] = self.head[v]
                else:
                    self.head[u] = u
                stack.append(u)

    def for_each(self, u, v):
        # [u, v]上の頂点集合の区間を列挙
        while True:
            if self.preorder[u] > self.preorder[v]:
                u, v = v, u
            l = max(self.preorder[self.head[v]], self.preorder[u])
            r = self.preorder[v]
            yield l, r # [l, r]
            if self.head[u] != self.head[v]:
                v = self.parent[self.head[v]]
            else:
                return

    def for_each_edge(self, u, v):
        # [u, v]上の辺集合の区間列挙
        # 辺の情報は子の頂点に
        while True:
            if self.preorder[u] > self.preorder[v]:
                u, v = v, u
            if self.head[u] != self.head[v]:
                yield self.preorder[self.head[v]], self.preorder[v]
                v = self.parent[self.head[v]]
            else:
                if u != v:
                    yield self.preorder[u]+1, self.preorder[v]
                break

    def subtree(self, v):
        # 頂点vの部分木の頂点集合の区間 [l, r)
        l = self.preorder[v]
        r = self.preorder[v]+self.size[v]
        return l, r

    def lca(self, u, v):
        # 頂点u, vのLCA
        while True:
            if self.preorder[u] > self.preorder[v]:
                u, v = v, u
            if self.head[u] == self.head[v]:
                return u
            v = self.parent[self.head[v]]

import sys
import io, os
input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline

n = int(input())
g = [[] for i in range(n)]
for i in range(n-1):
    a, b = map(int, input().split())
    a, b = a-1, b-1
    g[a].append(b)
    g[b].append(a)

INF = 10**18
from collections import deque
def bfs(s, g):
    n = len(g)
    d = [INF]*n
    d[s] = 0
    q = deque([])
    q.append(s)
    while q:
        v = q.popleft()
        for u in g[v]:
            if d[u] == INF:
                d[u] = d[v]+1
                q.append(u)
    return d

D1 = bfs(0, g)
M = max(D1)
for i in range(n):
    if D1[i] == M:
        x = i
        break
D2 = bfs(x, g)
M = max(D2)
for i in range(n):
    if D2[i] == M:
        y = i
        break

hld = HLD(g)
def dist(u, v):
    l = hld.lca(u, v)
    d = hld.depth[u]+hld.depth[v]-2*hld.depth[l]
    return d
d0 = dist(x,y)
res = 0
ans = []
for i in range(n):
    if i == x or i == y:
        continue
    d1 = dist(x, i)
    d2 = dist(y, i)
    temp = (d0+d1+d2)//2
    if temp > res:
        res = temp
        ans = [x+1, y+1, i+1]
print(res)
print(*ans)
