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) {
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt(), M = sc.nextInt();
        DJSet dj = new DJSet(N);
        for (int i = 0; i < M; i++) {
            dj.union(sc.nextInt()-1, sc.nextInt()-1);
        }
        System.out.println(dj.countSet()-1);
        sc.close();

    }

}

class DJSet {
    /**
     * 各要素の親を保持する配列。
     * 自信が根である要素は負の数を持つ。
     * （負の数の場合、絶対値がその集合の要素数）
     */
    public int[] upper;

    /**
     * コンストラクタ。
     * @param n 対象となる要素数
     */
    public DJSet(int n) {
        upper = new int[n];
        Arrays.fill(upper, -1);
    }

    /**
     * 要素xが含まれる集合のrootを求める。
     * rootを求める過程で経路圧縮しrootが親になるようにする。
     * @param x 要素x
     * @return 要素xが含まれる集合のroot
     */
    public int root(int x) {
        return upper[x] < 0 ? x : (upper[x] = root(upper[x]));
    }

    /**
     * 要素xと要素yが同じ集合か判定する。
     * @param x 要素x
     * @param y 要素y
     * @return 要素xと要素yが同じ集合であればtrue、そうでなければfalseを返す。
     */
    public boolean same(int x, int y) {
        return root(x) == root(y);
    }

    /**
     * 要素xと要素yを含むそれぞれの集合を合併する。
     * 要素数の多い集合のrootを残し、少ないほうのrootをその下に合併する。
     * @param x 要素x
     * @param y 要素y
     * @return 要素xと要素yを含むそれぞれの集合を合併できたらtrue、もともと同じ集合で合併しなかったらfalseを返す。
     */
    public boolean union(int x, int y) {
        x = root(x);
        y = root(y);
        if (x != y) {
            if (upper[y] < upper[x]) {
                int t = x;
                x = y;
                y = t;
            }
            upper[x] += upper[y];
            upper[y] = x;
        }
        return x != y;
    }

    /**
     * 集合の数を数える。
     * @return 集合の数
     */
    public int countSet() {
        int c = 0;
        for (int u : upper) {
            if (u < 0)
                c++;
        }
        return c;
    }

    /**
     * 要素ｘが含まれる集合の要素数を数える。
     * @param x 要素ｘ
     * @return 要素ｘが含まれる集合の要素数
     */
    public int countElement(int x) {
        return upper[root(x)] * -1;
    }
}