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.

Example

Input

3 1
1 2


Output

1
//package Atcoder;
import java.util.*;
import java.io.*;
public class Main
{
	static boolean vis[]=new boolean[(int)1e6+1];
	static final int mod=(int)1e9+7;
	static final int inf=(int)1e9;
	static final long INF=(long)1e18;
	public static void main(String[] args)
	{
		FastScanner fs=new FastScanner();
		int nodes=fs.nextInt();
		int edges=fs.nextInt();
		ArrayList<ArrayList<Integer>> adj = new ArrayList<ArrayList<Integer>>();
		for(int i=0;i<=nodes+1;i++)
		{
			adj.add(new ArrayList<Integer>());
		}
		for(int i=1;i<=edges;i++)
		{
			int u=fs.nextInt();
			int v=fs.nextInt();
			adj.get(u).add(v);
			adj.get(v).add(u);
		}
		int cc=0;
		int ans=0;
		for(int i=1;i<=nodes;i++)
		{
			if(!vis[i])
			{
				dfs(i,adj);
				cc++;
			}
		}
		ans+=(cc-1);
		for(int i=1;i<=nodes;i++)
		{
			if(!vis[i])
				ans++;
		}
		System.out.println(ans);
	}
	public static void dfs(int node,ArrayList<ArrayList<Integer>> adj)
	{
		vis[node]=true;
		for(int child:adj.get(node))
		{
			if(!vis[child])
				dfs(child,adj);
		}
	}
}
class FastScanner 
{
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer("");
		public String next() 
		{
			while (!st.hasMoreTokens())
				try {
					st=new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
			
		}
		
		long pow(long x,long n)
		{
			long res=1;
			while(n>0)
			{
			   	if(n%2!=0)
			   	{
			   		res=(res*x);n--;
			   	}
			   	else
			   	{
			   		x=(x*x);n/=2;
			   	}
			}
			return res;
		}
		int gcd(int a,int b)
		{
			if(b==0)
			{
				return a;
			}
			return gcd(b,a%b);
		}
		long longgcd(long a,long b)
		{
			if(b==0)
			{
				return a;
			}
			return longgcd(b,a%b);
		}
		int[] sort(int arr[])
		{
			ArrayList<Integer> list = new ArrayList<Integer>();
			for(int i:arr)list.add(i);
			Collections.sort(list);
			for(int i=0;i<arr.length;i++)
			{
				arr[i]=list.get(i);
			}
			return arr;
		}
		char[] charsort(char arr[])
	        {
	    	ArrayList<Character> list = new ArrayList<>();
	    	for(char c:arr)list.add(c);
	    	Collections.sort(list);
	    	for(int i=0;i<list.size();i++)
	    	{
	    		arr[i]=list.get(i);
	    	}
	    	return arr;
	        }
		long[] longsort(long arr[])
		{
			ArrayList<Long> list = new ArrayList<Long>();
			for(long i:arr)list.add(i);
			Collections.sort(list);
			for(int i=0;i<arr.length;i++)
			{
				arr[i]=list.get(i);
			}
			return arr;
		}
		public int nextInt() 
		{
			return Integer.parseInt(next());
		}
		public int[] readArray(int n)
		{
			int[] arr=new int[n];
			for (int i=0; i<n; i++) arr[i]=nextInt();
			return arr;
		}	
		public long nextLong()
		{
			return Long.parseLong(next());
		}
		public long[] longreadArray(int n) 
		{
			long[] a=new long[n];
			for (int i=0; i<n; i++) a[i]=nextLong();
			return a;
		}
}