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: Wrong Answer)

/*
 * main.c
 *
 *  Created on: 2020/03/28
 *      Author: family
 */

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#define MAX(a,b) (a > b ? a : b)
#define MIN(a,b) (a > b ? b : a)
typedef long long int ll;
typedef unsigned long long int ull;

int sort_inc(const void *a, const void *b) { return (*(int*)a - *(int*)b);}
int sort_dec(const void* a, const void* b) { return (*(int*)b - *(int*)a);}
int sort_dec_ll(const void *a, const void *b) {
    ll da = *(ll*)a, db = *(ll*)b;    int val = 0;
    if(da > db) { val = -1; }
    else if (da == db) { val = 0; }
    else { val = 1; }
    return val;
}
int sort_inc_ll(const void *a, const void *b) {
    ll da = *(ll*)a, db = *(ll*)b;    int val = 0;
    if(da > db) { val = 1; }
    else if (da == db) { val = 0; }
    else { val = -1; }
    return val;
}
int sort_dic(const void *a, const void *b) {
    char *pa = (char *)a;    char *pb = (char *)b;    int i = 0, val = 0, N = 10;
    for (i = 0; i < N; i++) {
    	char da = pa[i], db = pb[i];
    	if (da == db) continue;
  		if (da > db) val = 1; else val = -1;
        break;
    }
    return val;
}

void yesno(int ans) {
	if (ans == 1) {	printf("Yes\n");
	} else {		printf("No\n");	}
	return;
}

void okng(int ans) {
	if (ans == 1) { printf("OK\n");
	} else { 		printf("NG\n");	}
	return;
}

int AB[100001][2] = {0};
int city[100002] = {0};
int main()
{
	int N = 0, M = 0;
	scanf("%d %d", &N, &M);

	int i = 0;
	memset(AB, M, sizeof(int));
	for (i = 0; i < M; i++) {
		int a = 0, b = 0;
		scanf("%d %d", &a, &b);
		AB[i][0] = a;
		AB[i][1] = b;
		city[a]++;
		city[b]++;
	}

	int ans = 0;
	for (i = 1; i <= N; i++) {
		if (city[i] == 0) {
			ans++;
		}
	}

	printf("%d\n", ans);
	return 0;
}