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: Accepted)

'''
Created on 2020/09/26

@author: harurun
'''
#UnionFind(要素数)で初期化
class UnionFind():
    def __init__(self, n):
        self.n = n
        self.parents = [-1] * n
        #parents 各要素の親番号を格納するリストを返す

    #要素xが属するグループを返す
    def find(self, x):
        if self.parents[x] < 0:
            return x
        else:
            self.parents[x] = self.find(self.parents[x])
            return self.parents[x]
    #要素xとyが属するグループを併合する
    def union(self, x, y):
        x = self.find(x)
        y = self.find(y)

        if x == y:
            return

        if self.parents[x] > self.parents[y]:
            x, y = y, x

        self.parents[x] += self.parents[y]
        self.parents[y] = x

    #要素xが属するグループのサイズ
    def size(self, x):
        return -self.parents[self.find(x)]

    #要素xとyが同じグループに属するかどうか
    def same(self, x, y):
        return self.find(x) == self.find(y)

    #要素xが属するグループを返す
    def members(self, x):
        root = self.find(x)
        return [i for i in range(self.n) if self.find(i) == root]

    #すべての根の要素
    def roots(self):
        return [i for i, x in enumerate(self.parents) if x < 0]

    #グループの数
    def group_count(self):
        return len(self.roots())

    def all_group_members(self):
        return {r: self.members(r) for r in self.roots()}

    def __str__(self):
        return '\n'.join('{}: {}'.format(r, self.members(r)) for r in self.roots())

def main():
  import sys
  pin=sys.stdin.readline
  pout=sys.stdout.write
  perr=sys.stderr.write

  N,M=map(int,pin().split())
  ans=0
  uf=UnionFind(N)
  for i in range(M):
    A,B=map(int,pin().split())
    uf.union(A-1, B-1)
  print(uf.group_count()-1)
  return 
main()