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,n) for(ll i=0;i<(n);i++)
#define rrep(i,n) for(ll i = 1; i <= (n); ++i)
#define drep(i,n) for(ll i = (n)-1; i >= 0; --i)
#define all(v) v.begin(),v.end()
#define len(x) (ll)(x).length()
#define maxs(x,y) x = max(x,y)
#define mins(x,y) x = min(x,y)
#define pb push_back
#define sz(x) (ll)(x).size()
#define v(T) vector<T>
#define vv(T) vector<vector<T>>
using namespace std;
typedef long long ll;
typedef pair<ll,ll> P;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<ll> vl;
typedef vector<P> vp;
ll gcd(ll a,ll b){if(a%b==0){return b;}else{return(gcd(b,a%b));}}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
const int INF=1e9;
const ll MX = 1e18;
const ll MOD=INF+7;
const int di[] = {-1,0,1,0};
const int dj[] = {0,-1,0,1};
const double PI=acos(-1);
const string tr="abcdefghijklmnopqrstuvwxyz";
#define dame { puts("-1"); return 0;}
#define yn {puts("Yes");}else{puts("No");}
#define YN {puts("YES");}else{puts("NO");}
const ll MAX = 1000001;
long long fac[MAX], finv[MAX], inv[MAX];
void COMinit() {fac[0] = fac[1] = 1;finv[0] = finv[1] = 1;inv[1] = 1;for (ll i = 2; i < MAX; i++){fac[i] = fac[i - 1] * i % MOD;inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;finv[i] = finv[i - 1] * inv[i] % MOD;}}
long long COM(ll n, ll k){if (n < k) return 0;if (n < 0 || k < 0) return 0;return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;}
ll llpow(ll n,ll i){ll cnt=n;  for(ll j=0;j<i-1;j++){n*=cnt;}return n;}
bool ip/*is_prime*/(long long N) {if (N == 1) return false;for (long long i = 2; i * i <= N; ++i) {if (N % i == 0) return false;}return true;}
int digit(ll N) {int ans = 0;while (N) {++ans;N /= 10;}return ans;}
vector<pair<ll,ll>> pf/*prime_factorize*/(ll n){vector<pair<ll,ll>> res;for(ll a=2;a*a<=n;a++){if(n%a!=0) continue;ll ex=0;while(n%a==0){ex++;n/=a;}res.pb({a,ex});}if(n!=1) res.pb({n,1});return res;}
vector<ll> div/*divisor*/(ll n){  vector<ll> res={1};for(ll a=2;a*a<=n;a++){if(n%a!=0) continue;ll b=n/a;res.pb(b);if(b!=a) res.pb(a);}sort(all(res));return res;}

struct UnionFind {
    vector<int> par;

    UnionFind(int N) : par(N) {
        for(int i = 0; i < N; i++) par[i] = i;
    }

    int root(int x) {
        if (par[x] == x) return x;
        return par[x] = root(par[x]);
    }

    void unite(int x, int y) {
        int rx = root(x);
        int ry = root(y);
        if (rx == ry) return;
        par[rx] = ry;
    }

    bool same(int x, int y) {
        int rx = root(x);
        int ry = root(y);
        return rx == ry;
    }
};

int main() {
  int N, M;
  cin >> N >> M;

  UnionFind tree(N);

  for(int i = 0; i < M; i++) {
    int  x, y;
    cin >> x >> y;
    tree.unite(x-1, y-1);
  }

  int cnt=0;
  for(int i=0;i<N;i++){
    if(!tree.same(0,i)){
      tree.unite(0, i);
      cnt++;
    }
  }

  cout << cnt << endl;
  return 0;
}