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 Java (OpenJDK 11.0.6) (judge verdict: Accepted)

import java.util.*;

public class Main {
    public static void main(String[] args) throws Exception {
        // Your code here!
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int m = sc.nextInt();
        
        UnionFind uf = new UnionFind(n);
        while (m-- > 0) {
            int a = sc.nextInt()-1;
            int b = sc.nextInt()-1;
            
            uf.unite(a,b);
        }
        
        // 10こある
        // 連結成分8ko
        System.out.println(uf.connectedComponent-1);
    }
}

class UnionFind {
    int[] par;  // 自身の親
    int[] size; // ※(自身の属するグループの要素数はsize[uf.root(i)]。)
                // (※[i]全てに対して↑を更新するのは時間がかかるので)
    int connectedComponent;

    UnionFind(int n) {
        par = new int[n];
        for (int i = 0; i < n; i++) {
            par[i] = i;
        }
        
        size = new int[n];
        Arrays.fill(size, 1);
        connectedComponent = n;
    }

    int root(int x) {
        if (par[x] == x) return x;
        return par[x] = root(par[x]);
        // 1-2-4の場合、4の親=2を4の根=1に繋ぎ変えた上で親を返す
        // ↑根まで辿る深さ(計算量)を減らす為
        // return a = b;とは、aにbを代入し、aを返すという意味
    }
    
    // 点x, yを連結させる=同じグループにする
    void unite(int x, int y) {
        int rx = root(x);
        int ry = root(y);
        if (rx != ry) {
            par[ry] = rx;
            
            int resize = size[rx] + size[ry];
            size[rx] = resize;
            size[ry] = resize;
            connectedComponent--;
        }
    }
    
    // 同グループか否か
    boolean same(int x, int y) {
        int rx = root(x);
        int ry = root(y);
        return rx == ry;
    }
}