Ujan has a lot of useless stuff in his drawers, a considerable part of which are his math notebooks: it is time to sort them out. This time he found an old dusty graph theory notebook with a description of a graph.

It is an undirected weighted graph on n vertices. It is a complete graph: each pair of vertices is connected by an edge. The weight of each edge is either 0 or 1; exactly m edges have weight 1, and all others have weight 0.

Since Ujan doesn't really want to organize his notes, he decided to find the weight of the minimum spanning tree of the graph. (The weight of a spanning tree is the sum of all its edges.) Can you find the answer for Ujan so he stops procrastinating?

Input

The first line of the input contains two integers n and m (1 ≤ n ≤ 10^5, 0 ≤ m ≤ min((n(n-1))/(2),10^5)), the number of vertices and the number of edges of weight 1 in the graph. 

The i-th of the next m lines contains two integers a_i and b_i (1 ≤ a_i, b_i ≤ n, a_i ≠ b_i), the endpoints of the i-th edge of weight 1.

It is guaranteed that no edge appears twice in the input.

Output

Output a single integer, the weight of the minimum spanning tree of the graph.

Examples

Input


6 11
1 3
1 4
1 5
1 6
2 3
2 4
2 5
2 6
3 4
3 5
3 6


Output


2


Input


3 0


Output


0

Note

The graph from the first sample is shown below. Dashed edges have weight 0, other edges have weight 1. One of the minimum spanning trees is highlighted in orange and has total weight 2.

<image>

In the second sample, all edges have weight 0 so any spanning tree has total weight 0.
import java.util.*;
import java.io.*;
import java.text.*;
public class Main{
    //SOLUTION BEGIN
    //Into the Hardware Mode
    void pre() throws Exception{}
    void solve(int TC)throws Exception{
        int n = ni(), m = ni();
        int[] from = new int[m], to  =new int[m];
        for(int i = 0; i< m; i++){from[i] = ni()-1;to[i] = ni()-1;}
        int[][] g = makeU(n, from, to);
        int min = 0;
        int[] set = new int[n];
        for(int i = 0; i< n; i++)set[i] = i;
        for(int i = 0; i< n; i++)if(g[i].length < g[min].length)min = i;
        Arrays.sort(g[min]);
        int p = 0;
        for(int i = 0; i< n; i++){
            if(min == i)continue;
            if(p < g[min].length && g[min][p] == i)p++;
            else set[find(set, i)] = find(set, min);
        }
        for(int v:g[min]){
            Arrays.sort(g[v]);p = 0;
            for(int i = 0; i< n; i++){
                if(p < g[v].length && g[v][p] == i)p++;
                else set[find(set, i)] = find(set, v);
            }
        }
        int ans = -1;
        for(int i = 0; i< n; i++)if(find(set, i) == i)ans++;
        pn(ans);
    }
    int[][] makeU(int n, int[]from, int[] to){
        int[][] g = new int[n][];int[] cnt = new int[n];
        for(int i:from)cnt[i]++;
        for(int i:to)cnt[i]++;
        for(int i = 0; i< n; i++)g[i] = new int[cnt[i]];
        for(int i = 0; i< from.length; i++){
            g[from[i]][--cnt[from[i]]] = to[i];
            g[to[i]][--cnt[to[i]]] = from[i];
        }
        return g;
    }
    //SOLUTION END
    void hold(boolean b)throws Exception{if(!b)throw new Exception("Hold right there, Sparky!");}
    void exit(boolean b){if(!b)System.exit(0);}
    long IINF = (long)1e18;
    final int INF = (int)1e9, MX = (int)2e6+5;
    DecimalFormat df = new DecimalFormat("0.00");
    double PI = 3.141592653589793238462643383279502884197169399, eps = 1e-7;
    static boolean multipleTC = false, memory = false, fileIO = false;
    FastReader in;PrintWriter out;
    void run() throws Exception{
        if(fileIO){
            in = new FastReader("C:/users/user/desktop/inp.in");
            out = new PrintWriter("C:/users/user/desktop/out.out");
        }else {
            in = new FastReader();
            out = new PrintWriter(System.out);
        }
        //Solution Credits: Taranpreet Singh
        int T = (multipleTC)?ni():1;
        pre();
        for(int t = 1; t<= T; t++)solve(t);
        out.flush();
        out.close();
    }
    public static void main(String[] args) throws Exception{
        if(memory)new Thread(null, new Runnable() {public void run(){try{new Main().run();}catch(Exception e){e.printStackTrace();}}}, "1", 1 << 28).start();
        else new Main().run();
    }
    int find(int[] set, int u){return set[u] = (set[u] == u?u:find(set, set[u]));}
    int digit(long s){int ans = 0;while(s>0){s/=10;ans++;}return ans;}
    long gcd(long a, long b){return (b==0)?a:gcd(b,a%b);}
    int gcd(int a, int b){return (b==0)?a:gcd(b,a%b);}
    int bit(long n){return (n==0)?0:(1+bit(n&(n-1)));}
    void p(Object o){out.print(o);}
    void pn(Object o){out.println(o);}
    void pni(Object o){out.println(o);out.flush();}
    String n()throws Exception{return in.next();}
    String nln()throws Exception{return in.nextLine();}
    int ni()throws Exception{return Integer.parseInt(in.next());}
    long nl()throws Exception{return Long.parseLong(in.next());}
    double nd()throws Exception{return Double.parseDouble(in.next());}
 
    class FastReader{
        BufferedReader br;
        StringTokenizer st;
        public FastReader(){
            br = new BufferedReader(new InputStreamReader(System.in));
        }
 
        public FastReader(String s) throws Exception{
            br = new BufferedReader(new FileReader(s));
        }
 
        String next() throws Exception{
            while (st == null || !st.hasMoreElements()){
                try{
                    st = new StringTokenizer(br.readLine());
                }catch (IOException  e){
                    throw new Exception(e.toString());
                }
            }
            return st.nextToken();
        }
 
        String nextLine() throws Exception{
            String str = "";
            try{   
                str = br.readLine();
            }catch (IOException e){
                throw new Exception(e.toString());
            }  
            return str;
        }
    }   
}