Problem:
Score : 300 points

Problem Statement

There are N cities numbered 1 through N, and M bidirectional roads numbered 1 through M.
Road i connects City A_i and City B_i.

Snuke can perform the following operation zero or more times:

Choose two distinct cities that are not directly connected by a road, and build a new road between the two cities.

After he finishes the operations, it must be possible to travel from any city to any other cities by following roads (possibly multiple times).

What is the minimum number of roads he must build to achieve the goal?

Constraints

2 \leq N \leq 100,000

1 \leq M \leq 100,000

1 \leq A_i < B_i \leq N

No two roads connect the same pair of cities.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N M
A_1 B_1
:
A_M B_M

Output

Print the answer.

Sample Input 1

3 1
1 2

Sample Output 1

1

Initially, there are three cities, and there is a road between City 1 and City 2.

Snuke can achieve the goal by building one new road, for example, between City 1 and City 3.
After that,

We can travel between 1 and 2 directly.

We can travel between 1 and 3 directly.

We can travel between 2 and 3 by following both roads (2 - 1 - 3).

Solution in Python (3.8.2) (judge verdict: Runtime Error)

# Python3 program to print DFS traversal
# from a given given graph
from collections import defaultdict

viz = [0]*100005
# This class represents a directed graph using
# adjacency list representation


class Graph:

    # Constructor
    def __init__(self):

        # default dictionary to store graph
        self.graph = defaultdict(list)

    # function to add an edge to graph
    def addEdge(self, u, v):
        self.graph[u].append(v)

    # A function used by DFS
    def DFSUtil(self, v, visited):
        global viz
        # Mark the current node as visited
        # and print it
        visited[v] = True
        viz[v]=1
        # print(v, end=' ')

        # Recur for all the vertices
        # adjacent to this vertex
        for i in self.graph[v]:
            if visited[i] == False:
                self.DFSUtil(i, visited)

    # The function to do DFS traversal. It uses
    # recursive DFSUtil()
    def DFS(self, v,visited):

        # Mark all the vertices as not visited
        # visited = [False] * (max(self.graph)+1)

        # Call the recursive helper function
        # to print DFS traversal
        self.DFSUtil(v, visited)

        return visited

# Driver code


# Create a graph given
# in the above diagram

    
g = Graph()
n, m = map(int, input().split())
vizi=[False]*(n+1)
for i in range(m):
    a, b = map(int, input().split())
    g.addEdge(a, b)
    g.addEdge(b, a)

cnt = 0
vizi=g.DFS(0,vizi)
for i in range(1,n):
    if vizi[i] == False:
        cnt += 1
        vizi=g.DFS(i,vizi)

print(cnt)

# print("Following is DFS from (starting from vertex 2)")
# g.DFS(2)

# This code is contributed by Neelam Yadav