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.
def dfs(n, source, neighbors):
    # Distances à la racine
    dist = [-1] * n

    # Pointeur sur parent
    parent = [-1] * n

    # Hauteur du sous-arbre
    height = [-1] * n

    # Pointeur sur sous-arbre le plus haut
    height_ptr = [-1] * n

    # Sommet le plus éloigné
    deepest = 0
    deepest_ptr = source

    dist[source] = 0
    parent[source] = source

    stack = [source]

    while stack:
        u = stack[-1]
        is_processed = True

        for v in neighbors[u]:
            if parent[v] == -1:
                is_processed = False
                dist[v] = dist[u] + 1

                if dist[v] > deepest:
                    deepest = dist[v]
                    deepest_ptr = v

                parent[v] = u
                stack.append(v)

        if is_processed:
            stack.pop()

            max_height = -1
            max_height_ptr = -1

            for v in neighbors[u]:
                if parent[v] == u:
                    if height[v] > max_height:
                        max_height = height[v]
                        max_height_ptr = v

            height[u] = 1 + max_height
            height_ptr[u] = max_height_ptr

    return dist, parent, deepest, deepest_ptr, height, height_ptr

# Obtient les arêtes d’un diamètre d’un arbre et un arbre DFS issu de son
# premier sommet
def find_diameter(n, r, neighbors):
    _, _, _, v1, _, _ = dfs(n, r, neighbors)
    _, parent, size, v2, height, height_ptr = dfs(n, v1, neighbors)
    return v1, v2, size, height, height_ptr, parent

# Lecture de l’arbre
n = int(input())
neighbors = [[] for i in range(n)]

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

# Recherche du diamètre et soustraction de ses arêtes dans l’arbre
v1, v2, size, height, height_ptr, parent = find_diameter(n, 0, neighbors)

# Recherche d’une branche de taille maximale issue du diamètre
walker_prev = -1
walker = v2

max_height = 0
max_height_ptr = -1

while walker != v1:
    for branch in neighbors[walker]:
        if parent[branch] == walker and branch != walker_prev:
            if height[branch] + 1 > max_height:
                max_height = height[branch] + 1
                max_height_ptr = branch

    walker_prev = walker
    walker = parent[walker]

if max_height_ptr == -1:
    # Cas particulier : chemin
    print(size)
    print(v1 + 1, parent[v2] + 1, v2 + 1)

else:
    v3 = max_height_ptr

    while height_ptr[v3] != -1:
        v3 = height_ptr[v3]

    print(max_height + size)
    print(v1 + 1, v2 + 1, v3 + 1)
