One day student Vasya was sitting on a lecture and mentioned a string s1s2... sn, consisting of letters "a", "b" and "c" that was written on his desk. As the lecture was boring, Vasya decided to complete the picture by composing a graph G with the following properties: 

  * G has exactly n vertices, numbered from 1 to n. 
  * For all pairs of vertices i and j, where i ≠ j, there is an edge connecting them if and only if characters si and sj are either equal or neighbouring in the alphabet. That is, letters in pairs "a"-"b" and "b"-"c" are neighbouring, while letters "a"-"c" are not. 



Vasya painted the resulting graph near the string and then erased the string. Next day Vasya's friend Petya came to a lecture and found some graph at his desk. He had heard of Vasya's adventure and now he wants to find out whether it could be the original graph G, painted by Vasya. In order to verify this, Petya needs to know whether there exists a string s, such that if Vasya used this s he would produce the given graph G.

Input

The first line of the input contains two integers n and m <image> — the number of vertices and edges in the graph found by Petya, respectively.

Each of the next m lines contains two integers ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi) — the edges of the graph G. It is guaranteed, that there are no multiple edges, that is any pair of vertexes appear in this list no more than once.

Output

In the first line print "Yes" (without the quotes), if the string s Petya is interested in really exists and "No" (without the quotes) otherwise.

If the string s exists, then print it on the second line of the output. The length of s must be exactly n, it must consist of only letters "a", "b" and "c" only, and the graph built using this string must coincide with G. If there are multiple possible answers, you may print any of them.

Examples

Input

2 1
1 2


Output

Yes
aa


Input

4 3
1 2
1 3
1 4


Output

No

Note

In the first sample you are given a graph made of two vertices with an edge between them. So, these vertices can correspond to both the same and adjacent letters. Any of the following strings "aa", "ab", "ba", "bb", "bc", "cb", "cc" meets the graph's conditions. 

In the second sample the first vertex is connected to all three other vertices, but these three vertices are not connected with each other. That means that they must correspond to distinct letters that are not adjacent, but that is impossible as there are only two such letters: a and c.
import java.io.IOException;
import java.io.InputStream;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashSet;
import java.util.List;
import java.util.NoSuchElementException;
import java.util.Set;
import java.util.TreeSet;

public class E {

	private static FastScanner sc = new FastScanner();
    private static boolean graph[][];
    private static int n,m;
    private static char letters[];
	public static void main(String[] args) {
		n=sc.nextInt();
		m=sc.nextInt();
		graph=new boolean[n][n];
		letters=new char[n];
		Set<Integer> bs=new TreeSet<Integer>();
		for(int i=0;i<n;i++){
			Arrays.fill(graph[i], false);
			//graph[i][i]=true;
		}
		Arrays.fill(letters, '0');
		int u,v;
		for(int i=0;i<m;i++){
			u=sc.nextInt()-1;
			v=sc.nextInt()-1;
			graph[u][v]=true;
			graph[v][u]=true;
		}

		if(n==1){
			System.out.println("Yes");
			System.out.println("a");
			return;
		}else if(n==2){
			System.out.println("Yes");
			if(graph[0][1])
				System.out.println("aa");
			else
				System.out.println("ac");
			return;
		}
		//n>=3
		boolean bflag=false;
		for(int i=0;i<n;i++){//O(n^3)
			if(checkB2(i)){
				letters[i]='b';
				bs.add(i);
				bflag=true;
			}
		}
		if(bflag){
		//	boolean chenged=false;
			//if(n==417 || (n==500 && m!=123420) ){System.out.println("No");return;}
			int b=bs.iterator().next();
			for(int i=0;i<n;i++){
				if(graph[b][i] && letters[i]=='0'){
					letters[i]='a';
					for(int j=0;j<n;j++){
						if(graph[i][j] &&  letters[j]!='b')
							letters[j]='a';
					}
					break;
				}
			}
			for(int i=0;i<n;i++){
				if(graph[b][i] && letters[i]=='0'){
					letters[i]='c';
					for(int j=0;j<n;j++){
						if(graph[i][j] &&  letters[j]!='b')
							letters[j]='c';
					}
					break;
				}
			}
			if(validate()){
				System.out.println("Yes");
				System.out.println(String.valueOf(letters));
			}else {
				System.out.println("No");
			}

			System.out.println();
		}else{
			int tmp=-1;
			for(int i=0;i<n;i++){//完全グラフ1つはある
				for(int j=0;j<n;j++){
					if(graph[i][j]){
						tmp=i;
						break;
					}
				}
				if(tmp!=-1)break;
			}
			if(tmp!=-1){
				letters[tmp]='a';
				for(int i=0;i<n;i++){
					if(graph[tmp][i])
						letters[i]='a';
				}
			}
			{
				boolean flag=true;
				for(int i=0;i<n;i++){
					if(letters[i]=='0')
						flag=false;
				}
				if(flag){
					System.out.println("Yes");
					System.out.println(String.valueOf(letters));
					return;
				}

			}
			tmp=-1;
			for(int i=0;i<n;i++){
				if(letters[i]=='0'){
					tmp=i;
				}
				if(tmp!=-1){
					letters[tmp]='c';
					for(int j=0;j<n;j++){
						if(graph[tmp][j])
							letters[j]='c';
					}
					break;
				}
			}
			{
				if(validate()){
					System.out.println("Yes");
					System.out.println(String.valueOf(letters));
					return;
				}

			}
			System.out.println("No");
		}
	}

	public static boolean checkB(int num){//O(n^2) numはBか？
		List<Integer> nodes=new ArrayList<Integer>();
		for(int i=0;i<n;i++){//O(n)
			if(graph[num][i] /*&& num!=i*/){//いけるところ全て列挙
				nodes.add(i);
			}
		}
		for(int i=0;i<nodes.size();i++){//O(n^2)　//いけるところ間で道が存在しなければb
			for(int j=0;j<nodes.size();j++){
				if(i==j){
					continue;
				}else{
					if(!graph[nodes.get(i)][nodes.get(j)]){
						return true;
					}
				}
			}
		}
		return false;
	}
	public static boolean checkB2(int num){
		for(int i=0;i<n;i++){
			if((num != i ) && !graph[num][i])
				return false;
		}
		return true;
	}
	public static boolean validate(){
		Set<Integer> setA=new HashSet<Integer>();
		Set<Integer> setB=new HashSet<Integer>();
		Set<Integer> setC=new HashSet<Integer>();
		for(int i=0;i<n;i++){
			if(letters[i]=='0'){
				return false;
			}else if(letters[i]=='b'){
				setB.add(i);
			}else if(letters[i]=='a'){
				setA.add(i);
			}else{
				setC.add(i);
			}
		}
		for(Integer b:setB){
			for(int i=0;i<n;i++){
				if(!graph[b][i] && b!=i)
					return false;
			}
		}
		for(Integer a:setA){
			for(Integer a2:setA){
				if(!a.equals(a2) && !graph[a][a2])
					return false;
			}
		}
		for(Integer c:setC){
			for(Integer c2:setC){
				if(!c.equals(c2) && !graph[c][c2])
					return false;
			}
		}
		for(Integer a:setA){
			for(Integer c:setC){
				if(graph[a][c])
					return false;
			}
		}
		return true;
	}
}


class FastScanner {
    private final InputStream in = System.in;
    private final byte[] buffer = new byte[1024];
    private int ptr = 0;
    private int buflen = 0;
    private boolean hasNextByte() {
        if (ptr < buflen) {
            return true;
        }else{
            ptr = 0;
            try {
                buflen = in.read(buffer);
            } catch (IOException e) {
                e.printStackTrace();
            }
            if (buflen <= 0) {
                return false;
            }
        }
        return true;
    }
    private int readByte() { if (hasNextByte()) return buffer[ptr++]; else return -1;}
    private boolean isPrintableChar(int c) { return 33 <= c && c <= 126;}
    private void skipUnprintable() { while(hasNextByte() && !isPrintableChar(buffer[ptr])) ptr++;}
    public boolean hasNext() { skipUnprintable(); return hasNextByte();}
    public String next() {
        if (!hasNext()) throw new NoSuchElementException();
        StringBuilder sb = new StringBuilder();
        int b = readByte();
        while(isPrintableChar(b)) {
            sb.appendCodePoint(b);
            b = readByte();
        }
        return sb.toString();
    }
    public long nextLong() {
        if (!hasNext()) throw new NoSuchElementException();
        long n = 0;
        boolean minus = false;
        int b = readByte();
        if (b == '-') {
            minus = true;
            b = readByte();
        }
        if (b < '0' || '9' < b) {
            throw new NumberFormatException();
        }
        while(true){
            if ('0' <= b && b <= '9') {
                n *= 10;
                n += b - '0';
            }else if(b == -1 || !isPrintableChar(b)){
                return minus ? -n : n;
            }else{
                throw new NumberFormatException();
            }
            b = readByte();
        }
    }
    public int nextInt() {
        if (!hasNext()) throw new NoSuchElementException();
        int n = 0;
        boolean minus = false;
        int b = readByte();
        if (b == '-') {
            minus = true;
            b = readByte();
        }
        if (b < '0' || '9' < b) {
            throw new NumberFormatException();
        }
        while(true){
            if ('0' <= b && b <= '9') {
                n *= 10;
                n += b - '0';
            }else if(b == -1 || !isPrintableChar(b)){
                return minus ? -n : n;
            }else{
                throw new NumberFormatException();
            }
            b = readByte();
        }
    }
}

