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: Wrong Answer)


#include <bits/stdc++.h>
#define rep(i,n) for (ll i = 0; i < (n); ++i)
using namespace std;
using ll = long long;
using P = pair<ll,ll>;
const ll INF=1e18;

struct UnionFind {
  vector<int> d;
  UnionFind(int n): d(n,-1) {}//-1で初期化
  int root(int x) {
    if (d[x] < 0) return x;//rootを負にする（構成node数x(-1))
    //root以外は親の番号
    return d[x] = root(d[x]);
    //return(root(d[x]))とするとメモできないが上の書き方だとメモができる
  }
  bool unite(int x, int y) {
    x = root(x); y = root(y);
    if (x == y) return false;
    if (d[x] > d[y]) swap(x,y);
    //swapすることで小さい方を大きい方につけられ計算量減らせる
    d[x] += d[y];
    d[y] = x;
    return true;
  }
  bool same(int x, int y) { return root(x) == root(y);}
  //同じグループ化
  int size(int x) { return -d[root(x)];}
  //グループのnode数
    int group_num(int n){
        set<int> s;
        for(int i = 0; i < n; ++i){
            s.insert(root(i));
        }
        return int(s.size());}
};



int main() {
  ll n,m;
  cin>>n>>m;
  UnionFind uf(n);
  vector<ll> g(n);
  rep(i,m){
    ll a1,b1;
    cin>>a1>>b1;
    a1--;b1--;
    uf.unite(a1,b1);
  }
  ll now=uf.group_num(n);
  cout<<(n-now);
}