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.*;
import java.util.*;


public class Main {

    private static final double EPS = 1e-9;

    public static void main(String[] args) throws Exception{
        FastScanner fastScanner = new FastScanner(System.in);
        PrintWriter printWriter = new PrintWriter(System.out);
        Main main = new Main();
        main.solveC(fastScanner,printWriter);
        printWriter.close();

    }

    public void solveA(FastScanner scanner, PrintWriter writer){
        double d = scanner.nextDoube();
        double l = scanner.nextDoube();
        double v1 = scanner.nextDoube();
        double v2 = scanner.nextDoube();

        double ans = (l-d)/(v1+v2);

        writer.printf("%f\n",ans);

    }

    public void solveB(FastScanner scanner, PrintWriter writer){
        int n = scanner.nextInt();
        long[] nums = new long[n];
        for(int i=0;i<n;i++) nums[i]=scanner.nextLong();
        Arrays.sort(nums);
        HashMap<Long,Boolean> hm = new HashMap<>();
        long total = 0;
        for(int i=nums.length-1;i>=0;i--){
            long k = nums[i];
            while(k>0&&hm.containsKey(k)) k--;
            total+=k;
            hm.put(k,Boolean.TRUE);
        }

        writer.printf("%d\n",total);

    }
    int[] parent = null;

    int find(int i)
    {
        if(parent[i]==i) return parent[i];
        if (parent[i] == -1) {
            parent[i] = i;
            return parent[i];
        }
        parent[i]=find(parent[i]);
        return parent[i];
    }

    public void solveC(FastScanner scanner, PrintWriter writer){
        int n = scanner.nextInt();
        int m = scanner.nextInt();
        boolean[][] mat = new boolean[n][n];
        int[] cnt = new int[n];
        for(int i=0;i<m;i++){
            int a = scanner.nextInt();
            int b = scanner.nextInt();
            a--;
            b--;
            mat[a][b]=true;
            mat[b][a]=true;
            cnt[a]++;
            cnt[b]++;
        }
        char[] ansArr = new char[n];
        Arrays.fill(ansArr,'1');
        for(int i=0;i<n;i++){
            if(cnt[i]==n-1) {
                ansArr[i] = 'b';
                for(int j=0;j<n;j++){
                    mat[i][j]=false;
                    mat[j][i]=false;
                }
            }
        }
        for(int i=0;i<n;i++){
            if(ansArr[i]=='1'){
                ansArr[i]='a';
                for(int j=0;j<n;j++){
                    if(mat[i][j])
                        ansArr[j]='a';
                }
                break;
            }
        }

        for(int i=0;i<n;i++){
            if(ansArr[i]=='1'){
                ansArr[i]='c';
                for(int j=0;j<n;j++){
                    if(mat[i][j]){
                        if(ansArr[j]=='1')
                            ansArr[j]='c';
                        else{
                            writer.printf("No\n");
                            return;
                        }
                    }

                }
                break;
            }
        }

        for(int i=0;i<n;i++){
            if(ansArr[i]=='1'){
                writer.printf("No\n");
                return;
            }
        }

        for(int i=0;i<n;i++){
            if(ansArr[i]=='a'){
                for(int j=i+1;j<n;j++){
                    if((ansArr[j]=='a'&&mat[i][j]==false)||(ansArr[j]=='c'&&mat[i][j]==true)){
                        writer.printf("No\n");
                        return;
                    }
                }
            }

            if(ansArr[i]=='c'){
                for(int j=i+1;j<n;j++){
                    if((ansArr[j]=='c'&&mat[i][j]==false)||(ansArr[j]=='a'&&mat[i][j]==true)){
                        writer.printf("No\n");
                        return;
                    }
                }
            }


        }

        writer.printf("Yes\n%s\n",new String(ansArr));

    }

    static class Point{
        long x,y;
        public Point(long x, long y){
            this.x = x;
            this.y = y;
        }
    }

    long doubleAreaOfTriangle(Point a, Point b, Point c) {
        long ar = a.x*(b.y-c.y)+b.x*(c.y-a.y)+c.x*(a.y-b.y);
        if(ar<0) ar=ar*(-1);
        return ar;
    }

    boolean isCisOnAB(Point a, Point b, Point c){
        long cross =  (c.y - a.y) * (b.x - a.x) - (c.x - a.x) * (b.y - a.y);
        if(cross!=0) return false;
        long dotproduct = (c.x - a.x) * (b.x - a.x) + (c.y - a.y)*(b.y - a.y);
        if(dotproduct < 0)
            return false;
        long squaredlengthba = (b.x - a.x)*(b.x - a.x) + (b.y - a.y)*(b.y - a.y);
        if (dotproduct > squaredlengthba) return false;

        return true;

    }



    static class FastScanner {
        private BufferedReader reader = null;
        private StringTokenizer tokenizer = null;

        public FastScanner(InputStream in) {
            reader = new BufferedReader(new InputStreamReader(in));
            tokenizer = null;
        }

        public String next() {
            if (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                    return next();
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

        public String nextLine() {
            if (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    return reader.readLine();
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }

            return tokenizer.nextToken("\n");
        }

        public long nextLong() {
            return Long.parseLong(next());
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }

        public double nextDoube()  {
            return Double.parseDouble(next());
        }

    }


}
