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: Wrong Answer)

class UnionFind:
    __slots__ = ["N", "root"]

    def __init__(self, N):
        """
        N:要素数
        root:各要素の親要素の番号を格納するリスト.
             ただし, root[x] < 0 ならその頂点が根で-root[x]が木の要素数.
        """
        self.N = N
        self.root = [-1] * N

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

    def find(self, x):
        """頂点xの根を見つける"""
        r = x
        while self.root[r] >= 0:
            r = self.root[r]

        while self.root[x] >= 0:
            self.root[x], x = r, self.root[x]

        return r

    def union(self, x, y):
        """x, yが属する木をunion"""
        x = self.find(x)
        y = self.find(y)
        if x == y:
            return
        elif self.root[y] < self.root[x]:
            x, y = y, x
        self.root[x] += self.root[y]
        self.root[y] = x

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

    def count(self, x):
        """頂点xが属する木のサイズを返す"""
        return - self.root[self.find(x)]

    def members(self, x):
        """xが属する木の要素を列挙"""
        _root = self.find(x)
        return [i for i in range(self.N) if self.find(i) == _root]

    def roots(self):
        """森の根を列挙"""
        return [i for i, x in enumerate(self.root) if x < 0]

    def group_count(self):
        """連結成分の数"""
        return len(self.roots())

    def all_group_members(self):
        """{ルート要素: [そのグループに含まれる要素のリスト], ...}の辞書を返す"""
        return {r: self.members(r) for r in self.roots()}


import sys
sys.setrecursionlimit(1000000000)
input = sys.stdin.readline
n, m = map(int, input().split())
uf = UnionFind(n)
for i in range(m):
    x, y = map(int, input().split())
    uf.union(x - 1, y - 1)
k = uf.group_count()
print(n - k)