I found an interesting problem on https://codeforces.com/problemset/problem/1029/E:
You are given an undirected tree consisting of $n$ vertices. An undirected tree is a connected undirected graph with $n - 1$ edges.

Your task is to add the minimum number of edges in such a way that the length of the shortest path from the vertex $1$ to any other vertex is at most $2$. Note that you are not allowed to add loops and multiple edges.


-----Input-----

The first line contains one integer $n$ ($2 \le n \le 2 \cdot 10^5$) — the number of vertices in the tree.

The following $n - 1$ lines contain edges: edge $i$ is given as a pair of vertices $u_i, v_i$ ($1 \le u_i, v_i \le n$). It is guaranteed that the given edges form a tree. It is guaranteed that there are no loops and multiple edges in the given edges.


-----Output-----

Print a single integer — the minimum number of edges you have to add in order to make the shortest distance from the vertex $1$ to any other vertex at most $2$. Note that you are not allowed to add loops and multiple edges.


-----Examples-----
Input
7
1 2
2 3
2 4
4 5
4 6
5 7

Output
2

Input
7
1 2
1 3
2 4
2 5
3 6
1 7

Output
0

Input
7
1 2
2 3
3 4
3 5
3 6
3 7

Output
1



-----Note-----

The tree corresponding to the first example: [Image] The answer is $2$, some of the possible answers are the following: $[(1, 5), (1, 6)]$, $[(1, 4), (1, 7)]$, $[(1, 6), (1, 7)]$.

The tree corresponding to the second example: [Image] The answer is $0$.

The tree corresponding to the third example: [Image] The answer is $1$, only one possible way to reach it is to add the edge $(1, 3)$.

I tried it in Python, but could not do it. Can you solve it?
import sys

def get_new_edges(graph):
    n = len(graph)
    far_vertex = []
    pi = [None]*n
    visit = [False]*n
    visit[0]
    queue = [[0,0]]
    i = 0    
    while True:
        if i >= len(queue): break
        current, d = queue[i]
        i += 1
        visit[current] = True
        for v in graph[current]:
            if not visit[v]:
                u = [v, d+1]
                pi[v] = current
                queue.append(u)
                if d+1 > 2:
                    far_vertex.append(u)
    
    far_vertex.sort(key=lambda x: -x[1])

    pos = [None]*n
    for i, e in enumerate(far_vertex):
        pos[e[0]] = i

    count = 0
    for i in range(len(far_vertex)):
        if not far_vertex[i]: continue
        vertex, depth = far_vertex[i]
        father = pi[vertex]
        count += 1
        if pos[father]:
            far_vertex[pos[father]] = None
        for u in graph[father]:
            if pos[u]:
                far_vertex[pos[u]] = None

    return count
    
def read_int_line():
    return list(map(int, sys.stdin.readline().split()))

vertex_count = int(input())
graph = [[] for _ in range(vertex_count)]

for i in range(vertex_count - 1):
    v1, v2 = read_int_line()
    v1 -= 1
    v2 -= 1
    graph[v1].append(v2)
    graph[v2].append(v1)

print(get_new_edges(graph))