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)

import os, sys
from io import IOBase, BytesIO
py2 = round(0.5)
if py2:
    from future_builtins import ascii, filter, hex, map, oct, zip
    range = xrange
BUFSIZE = 8192
class FastIO(BytesIO):
    newlines = 0
 
    def __init__(self, file):
        self._file = file
        self._fd = file.fileno()
        self.writable = 'x' in file.mode or 'w' in file.mode
        self.write = super(FastIO, self).write if self.writable else None
 
    def _fill(self):
        s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
        self.seek((self.tell(), self.seek(0,2), super(FastIO, self).write(s))[0])
        return s
 
    def read(self):
        while self._fill(): pass
        return super(FastIO,self).read()
 
    def readline(self):
        while self.newlines == 0:
            s = self._fill(); self.newlines = s.count(b'\n') + (not s)
        self.newlines -= 1
        return super(FastIO, self).readline()
 
    def flush(self):
        if self.writable:
            os.write(self._fd, self.getvalue())
            self.truncate(0), self.seek(0)
 
class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        if py2:
            self.write = self.buffer.write
            self.read = self.buffer.read
            self.readline = self.buffer.readline
        else:
            self.write = lambda s:self.buffer.write(s.encode('ascii'))
            self.read = lambda:self.buffer.read().decode('ascii')
            self.readline = lambda:self.buffer.readline().decode('ascii')
 
 
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip('\r\n')
 
# Cout implemented in Python
import sys
class ostream:
    def __lshift__(self,a):
        sys.stdout.write(str(a))
        return self
cout = ostream()
endl = '\n'




def solve():
    n, m = map(int, input().split())

    adj_list = {x: [] for x in range(1, n + 1)}

    for _ in range(m):
        a, b = map(int, input().split())
        adj_list[a].append(b)
        adj_list[b].append(a)

    sets = 0
    visited = {}

    def bfs(curr_node):
        nonlocal visited

        q = [curr_node]

        while q:
            ele = q.pop(0)
            visited[ele] = True
            
            for i in adj_list[curr_node]:
                if i not in visited:
                    q.append(i)
    
    for i in range(1, n + 1):
        if i not in visited:
            sets += 1
            bfs(i)

    cout<<sets - 1<<endl




def main():
    solve()


if __name__ == "__main__":
    main()