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)

import sys
from collections import deque, defaultdict, Counter
from itertools import accumulate, product, permutations, combinations
from operator import itemgetter
from bisect import bisect_left, bisect_right
from heapq import heappop, heappush
from math import ceil, floor, sqrt, gcd, inf
from copy import deepcopy
import numpy as np
import scipy as sp

class UnionFind():
    def __init__(self, n):
        self.n = n
        self.parents = [-1 for i in range(n)]

    def find(self, x):
        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):
        x = self.find(x)
        y = self.find(y)
        if x != y:
            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):
        return -self.parents[self.find(x)]

    def same(self, x, y):
        return self.find(x) == self.find(y)

    def members(self, x):
        root = self.find(x)
        return [i for i in range(self.n) if self.find(i) == root]

    @property
    def roots(self):
        return [i for i, x in enumerate(self.parents) if x < 0]

    @property
    def group_count(self):
        return len(self.roots)

    @property
    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)

INF = inf
MOD = 1000000007

n, m = [int(i) for i in input().split()]
A = [[int(i) for i in input().split()]for j in range(m)]    # nは行数

tmp = 0
res = 0

uf = UnionFind(n)
for i in range(m):
    a = A[i][0] - 1
    b = A[i][1] - 1
    uf.union(a, b)
res = uf.group_count - 1

print(res)