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)

class UnionFind():
    def __init__(self, n):
        self.n = n
        self.parents = [-1] * n

    def find(self, x):
        """要素xが属するグループの根を返す"""
        if self.parents[x] < 0:
            return x
        else:
            self.parents[x] = self.find(self.parents[x])
            return self.parents[x]

    def union(self, x, y):
        """要素xが属するグループと要素yが属するグループを併合"""
        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

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

    def same(self, x, y):
        """要素x, yが同じグループかどうか"""
        return self.find(x) == self.find(y)

    def members(self, x):
        """要素xが属するグループの要素をリストで返す"""
        root = self.find(x)
        return [i for i in range(self.n) if self.find(i) == root]

    def roots(self):
        """すべての根の要素をリストで返す"""
        return [i for i, x in enumerate(self.parents) if x < 0]

    def group_count(self):
        """グループの数を返す"""
        return len(self.roots())

    def all_group_members(self):
        """{ルート要素: [そのグループの要素, ...], ...}を返す"""
        return {r: self.members(r) for r in self.roots()}

    def __str__(self):
        """print()での表示用"""
        return '\n'.join('{}: {}'.format(r, self.members(r)) for r in self.roots())

N, M = map(int, input().split())
uf = UnionFind(N)
for _ in range(M):
    a, b = map(int, input().split())
    a, b = a - 1, b - 1
    uf.union(a, b)

print(uf.group_count() - 1)