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)

#include<bits/stdc++.h>

#define rep(i,a,n)     for(int i=a;i<n;i++)
#define rep3(n)        for(int i=0;i<n;i++)
#define ll             long long
#define int            long long
#define pb             push_back
#define all(v)         v.begin(),v.end()
#define endl           "\n"
#define eb             emplace_back
#define x              first
#define y              second
#define gcd(a,b)       __gcd(a,b)
#define mem1(a)        memset(a,-1,sizeof(a))
#define mem0(a)        memset(a,0,sizeof(a))
#define pres(a,x)      a.find(x)!=a.end()
#define sz(a)          (int)a.size()
#define pii            pair<int,int>
#define hell           1000000007
#define elasped_time   1.0 * clock() / CLOCKS_PER_SEC
using namespace std;


template<typename T1,typename T2>istream& operator>>(istream& in,pair<T1,T2> &a){in>>a.x>>a.y;return in;}
template<typename T1,typename T2>ostream& operator<<(ostream& out,pair<T1,T2> a){out<<a.x<<" "<<a.y;return out;}
template<typename T,typename T1>T maxs(T &a,T1 b){if(b>a)a=b;return a;}
template<typename T,typename T1>T mins(T &a,T1 b){if(b<a)a=b;return a;}

const int maxn=2000005;
int p[maxn];
int sz[maxn];
int cnt;
void clear(int n=maxn){
    rep(i,0,maxn)p[i]=i,sz[i]=1;
}
int root(int x){
   while(x!=p[x]){
       p[x]=p[p[x]];
       x=p[x];
   }
   return x;  
}
void merge(int x,int y){
    int p1=root(x);
    int p2=root(y);
    if(p1==p2)return;
    cnt--;
    if(sz[p1]>=sz[p2]){
        p[p2]=p1;
        sz[p1]+=sz[p2];
    }
    else{
        p[p1]=p2;
        sz[p2]+=sz[p1];
    }
}

int solve(){
 	int n,m;cin>>n>>m;
 	clear();
 	cnt=n;
 	rep3(m){
 		int a,b;cin>>a>>b;
 		merge(a,b);
 	}
 	cout<<cnt-1<<endl;
 return 0;
}
signed main(){
    ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
    //freopen("input.txt", "r", stdin);
    //freopen("output.txt", "w", stdout);
    #ifdef SIEVE
    sieve();
    #endif
    #ifdef NCR
    init();
    #endif
    int t=1;//cin>>t;
    while(t--){
        solve();
    }
    return 0;
}