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 C++ (GCC 9.2.1) (judge verdict: Accepted)

/*  -*- coding: utf-8 -*-
 *
 * c.cc: C - Connect Cities
 */

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<iostream>
#include<string>
#include<vector>
#include<map>
#include<set>
#include<stack>
#include<list>
#include<queue>
#include<deque>
#include<algorithm>
#include<numeric>
#include<utility>
#include<complex>
#include<functional>

using namespace std;

/* constant */

const int MAX_N = 100000;

/* typedef */

struct UFT {
  int links[MAX_N], ranks[MAX_N], sizes[MAX_N];
  UFT() {}

  void init(int n) {
    for (int i = 0; i < n; i++) links[i] = i, ranks[i] = sizes[i] = 1;
  }

  int root(int i) {
    int i0 = i;
    while (links[i0] != i0) i0 = links[i0];
    return (links[i] = i0);
  }

  int rank(int i) { return ranks[root(i)]; }
  int size(int i) { return sizes[root(i)]; }
  bool same(int i, int j) { return root(i) == root(j); }

  int merge(int i0, int i1) {
    int r0 = root(i0), r1 = root(i1), mr;
    if (r0 == r1) return r0;
    if (ranks[r0] == ranks[r1]) {
      links[r1] = r0;
      sizes[r0] += sizes[r1];
      ranks[r0]++;
      mr = r0;
    }
    else if (ranks[r0] > ranks[r1]) {
      links[r1] = r0;
      sizes[r0] += sizes[r1];
      mr = r0;
    }
    else {
      links[r0] = r1;
      sizes[r1] += sizes[r0];
      mr = r1;
    }
    return mr;
  }
};

/* global variables */

UFT uft;
int gs[MAX_N];

/* subroutines */

/* main */

int main() {
  int n, m;
  scanf("%d%d", &n, &m);

  uft.init(n);

  for (int i = 0; i < m; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    a--, b--;
    uft.merge(a, b);
  }

  memset(gs, -1, sizeof(gs));
  int k = 0;

  for (int i = 0; i < n; i++) {
    int ri = uft.root(i);
    if (gs[ri] < 0) gs[ri] = k++;
  }

  printf("%d\n", k - 1);
  return 0;
}