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)

//#define _GLIBCXX_DEBUG
#include <bits/stdc++.h>
#define rep(i, n) for(int i=0; i<n; ++i)
#define all(v) v.begin(), v.end()
#define rall(v) v.rbegin(), v.rend()
using namespace std;
using ll = int64_t;
using ld = long double;
using P = pair<int, int>;
using vs = vector<string>;
using vi = vector<int>;
using vvi = vector<vi>;
template<class T> using PQ = priority_queue<T>;
template<class T> using PQG = priority_queue<T, vector<T>, greater<T> >;
const int INF = 0xccccccc;
const ll LINF = 0xcccccccccccccccLL;
template<typename T1, typename T2>
inline bool chmax(T1 &a, T2 b) {return a < b && (a = b, true);}
template<typename T1, typename T2>
inline bool chmin(T1 &a, T2 b) {return a > b && (a = b, true);}
template<typename T1, typename T2>
istream &operator>>(istream &is, pair<T1, T2> &p) { return is >> p.first >> p.second;}
template<typename T1, typename T2>
ostream &operator<<(ostream &os, const pair<T1, T2> &p) { return os << p.first << ' ' << p.second;}

struct UnionFind {
  vector<int> par; // 親を指すvector,-par[親]は木のサイズ
  UnionFind(int n):par(n, -1) {} // uniteで親を埋め込んでいく必要あり
  int root(int x) { // 親をたどる&データの整理
    if(par[x] < 0) return x;
    return par[x] = root(par[x]);
  }
  bool unite(int x, int y) { // データの結合
    x = root(x);
    y = root(y);
    if(x == y) return false;
    if(par[x] > par[y]) swap(x, y);
    par[x] += par[y];
    par[y] = x;
    return true;
  }
  bool same(int x, int y) {return root(x) == root(y);} // 所属判定
  int size(int x) {return -par[root(x)];} // 木のサイズ
};

//head

int n, m;
int ans;

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m;
  UnionFind uft(n);
  rep(i, m) {
    int a, b;
    cin >> a >> b;
    a--; b--;
    uft.unite(a, b);
  }
  rep(i, n) if(uft.root(i) == i) ans++;
  cout << ans-1 << endl;
}