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 PyPy3 (7.3.0) (judge verdict: Accepted)

# coding: utf-8
N, Q = map(int,input().split())
#N,Q=map(int,input().split())
#親のノードを格納　2の親が4のとき par[2]=4
par=[0]
#木の深さを格納　根が2の深さが3の時 rank[2]=3
rank=[0]

#初期化
for i in range(1,N+1):
    par.append(i)
    rank.append(0)

#xの根を探す、同時に通過したノードすべてを根に直接つける
def find(x):
    #xの親がxならば根なのでxを返す
    if par[x]==x:
        return x
    else:
        #違う場合は親の親を捜し自分の親にする
        par[x] = find(par[x])
        #再帰的に行うと根が見つかる
        return par[x]

#xが属する木とyが属する木を併合する
def unite(x,y):
    #根を探す
    root_x=find(x)
    root_y=find(y)

    #根が同じなら元からつながっているので終わり
    if root_x == root_y:
        return

    #ランクが大きい木の根の直下にランクが小さい木を結合
    if rank[root_x] < rank[root_y]:
        par[root_x] = root_y

    else:
        par[root_y] = root_x

    #ランクが等しいときのみランクが増える#なぜ#yにxが結合？
    if rank[root_x]==rank[root_y]:
        rank[root_x]+=1

#xとyが同じ木に存在するかを調べる
def same(x,y):
    return find(x)==find(y)

#main
for i in range(Q):
    a,b=map(int,input().split())
    unite(a, b)
res = []
for i in range(1, N+1):
    res.append(find(i))
ans = len(set(res))-1
print(ans)