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)

/*starting with the name of almighty ALLAH*/
/*solved by RAKIB JOY*/
// Practice is the only shortcut to improve
#include<bits/stdc++.h>
#define ll long long int
#define pb push_back
#define vc vector
#define vi vc<int>
#define vl vc<ll>
#define dbg(x) cout<<x<<endl;
#define mp(x,y) make_pair(x,y)
#define yes dbg("YES")
#define no dbg("NO")
#define tst int t;cin>>t;while(t--)
#define srt(v) sort(v.begin(),v.end());
#define rsrt(v) sort(v.rbegin(),v.rend());
#define rj ios::sync_with_stdio(false);cin.tie(0);
#define rvs(v) reverse(v.begin(),v.end());
#define F first
#define S second
#define mod 1000000007
#define gcd(a,b) __gcd(a,b)
#define lcm(a,b) (a*b)/gcd(a,b)
#define PI 2*acos(0.0)
#define pii pair<int,int>
#define fr(i,a,b) for(ll i=a;i<=b;i++)
#define coutv(v) for(auto it:v)cout<<it<<' ';cout<<endl;
#define cinv(v) for(auto &it:v)cin>>it;
#define all(v) v.begin(),v.end()
#define rall(v) v.rbegin(),v.rend()
using namespace std;
/*#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
template<typename T> using ordered_set =tree<T, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update>;
*/
//sieve bool b[1000005];void primenire(){int i, j;b[0]=true;b[1]=true;for(i=4; i<=1000000; i+=2){b[i]=true;}for(i=3; i<=1001; i+=2){if(!b[i]){for(j=i*i; j<=1000001; j+=2*i){b[j]=true;}}}}
// power ll power(ll a,ll b){if(b==0)return 1LL;if(b&1)return a*power(a,b-1);else{ll x=power(a,b/2);return x*x;}}
//bigmod long long int m(long long int a,long long int b,long long int c){if(b==0)return 1;if(b%2==0){long long int  tmp=m(a,b/2,c);return ((tmp%c)*(tmp%c))%c;}elsereturn ((a%c)*(m(a,b-1,c)%c))%c;}
//MaxSubarraySum ll mss(vector<ll>v){ll mx=v[0];ll sz=v.size();for(int i=1;i<=sz;i++){ll t=v[i]+v[i-1];if(t>=v[i]) v[i]=t;mx=max(mx,v[i]*1LL);}return mx;}
//combination ll com[1001][1001];void combination(){com[1][1]=1;com[1][0]=1;for(ll i=2; i<=1000; i++){com[i][0]=1;for(ll j=1; j<=i; j++){com[i][j]=com[i-1][j]+com[i-1][j-1];}}}
///--------------Graph Moves--------------------------------------
///const int fx[] = {+1,-1,+0,+0};
///const int fy[] = {+0,+0,+1,-1};
///const int fx[] = {+0,+0,+1,-1,-1,+1,-1,+1}; ///King's move
///const int fy[] = {-1,+1,+0,+0,+1,+1,-1,-1}; ///king's Move
///const int fx[] = {-2,-2,-1,-1,+1,+1,+2,+2}; ///knight's move
///const int fy[] = {-1,+1,-2,+2,-2,+2,-1,+1}; ///knight's move
///---------------------------------------------------------------
vi v[100005];
bool vis[100005];
void dfs(ll n)
{
   if(vis[n]) return;
   vis[n]=1;
   for(auto it:v[n]) dfs(it);
}
void solve()
{
    ll a,b;
    cin>>a>>b;
    fr(i,1,b)
    {
       ll x,y;
       cin>>x>>y;
       v[x].pb(y);
       v[y].pb(x);
    }
    ll cnt=0;
    fr(i,1,a)
    {
       if(!vis[i])
       {
          dfs(i);
          cnt++;
       }
    }
    cout<<cnt-1<<endl;
}
int main()
{
    rj
    solve();
    return 0;
}