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>
using namespace std;
#define ll long long
#define n_l '\n'

#ifndef ONLINE_JUDGE
#define dbg(...) cout << "[" << #__VA_ARGS__ << "]: "; cout << to_string(__VA_ARGS__) << endl
template <typename T, size_t N> int SIZE(const T (&t)[N]){ return N; } template<typename T> int SIZE(const T &t){ return t.size(); } string to_string(const string s, int x1=0, int x2=1e9){ return '"' + ((x1 < s.size()) ? s.substr(x1, x2-x1+1) : "") + '"'; } string to_string(const char* s) { return to_string((string) s); } string to_string(const bool b) { return (b ? "true" : "false"); } string to_string(const char c){ return string({c}); } template<size_t N> string to_string(const bitset<N> &b, int x1=0, int x2=1e9){ string t = ""; for(int __iii__ = min(x1,SIZE(b)),  __jjj__ = min(x2, SIZE(b)-1); __iii__ <= __jjj__; ++__iii__){ t += b[__iii__] + '0'; } return '"' + t + '"'; } template <typename A, typename... C> string to_string(const A (&v), int x1=0, int x2=1e9, C... coords); int l_v_l_v_l = 0, t_a_b_s = 0; template <typename A, typename B> string to_string(const pair<A, B> &p) { l_v_l_v_l++; string res = "(" + to_string(p.first) + ", " + to_string(p.second) + ")"; l_v_l_v_l--; return res; } template <typename A, typename... C> string to_string(const A (&v), int x1, int x2, C... coords) { int rnk = rank<A>::value; string tab(t_a_b_s, ' '); string res = ""; bool first = true; if(l_v_l_v_l == 0) res += n_l; res += tab + "["; x1 = min(x1, SIZE(v)), x2 = min(x2, SIZE(v)); auto l = begin(v); advance(l, x1); auto r = l; advance(r, (x2-x1) + (x2 < SIZE(v))); for (auto e = l; e != r; e = next(e)) { if (!first) { res += ", "; } first = false; l_v_l_v_l++; if(e != l){ if(rnk > 1) { res += n_l; t_a_b_s = l_v_l_v_l; }; } else{ t_a_b_s = 0; } res += to_string(*e, coords...); l_v_l_v_l--; } res += "]"; if(l_v_l_v_l == 0) res += n_l; return res; } void dbgm(){;} template<typename Heads, typename... Tails> void dbgm(Heads H, Tails... T){ cout << to_string(H) << " | "; dbgm(T...); } 
#define dbgm(...) cout << "[" << #__VA_ARGS__ << "]: "; dbgm(__VA_ARGS__); cout << endl
#endif

#define pb push_back
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define sz(x) ((int)(x).size())
#define forn(i,n) for(int i=0;i<n;i++)
#define for1(i,n) for(int i=1;i<=n;i++)
typedef vector<int> vi;
typedef vector<ll> vl;
typedef pair<int,int> pii;
typedef double db;
typedef vector<vi> vvi;
const ll mod=1000000007;
const ll N=100005; //check this
int dx[] = {-1, 0, 1, 0, 1, 1, -1, -1};
int dy[] = {0, -1, 0, 1, 1, -1, 1, -1};

ll powmod(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll powb(ll a,ll b) {ll res=1;a ; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a;a=a*a;}return res;}
ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;}

vi g[N];
int vis[N];
void dfs(int s){
    vis[s]=1;
    for(auto i:g[s]){
        if(!vis[i])dfs(i);
    }
}

int main(){
    ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);

    #ifndef ONLINE_JUDGE
    freopen("inputf.in", "r", stdin);
    freopen("outputf.in", "w", stdout);
    #endif

    int t=1;
    //cin>>t;
    while(t--){
        int n,m;cin>>n>>m;
        forn(i,m){
            int x,y;cin>>x>>y;
            g[x].pb(y);
            g[y].pb(x);
        }
        int ans=0;
        for1(i,n){
            if(!vis[i]){
                ans++;
                dfs(i);
            }
        }
        cout<<ans-1;
    }
}