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.util.ArrayList;
import java.util.InputMismatchException;
import java.util.LinkedList;
import java.util.Scanner;

public class q3 
{
	static node g[];
	public static void main(String args[])
	{
		FasterScanner s=new FasterScanner();
		int n=s.nextInt();
		int m=s.nextInt();
		LinkedList<node> l=new LinkedList<q3.node>();
		g=new node[n];
		for(int i=0;i<n;i++)
			{
			g[i]=new node(i);
			
			}
		for(int i=0;i<m;i++)
		{
			int a=s.nextInt()-1;
			int b=s.nextInt()-1;
			g[a].adj.add(g[b]);
			g[b].adj.add(g[a]);
			g[a].d++;
			g[b].d++;	
		}
		int curr=n-1;
		for(int i=0;i<n;i++)
		{
			if(g[i].d==n-1)
			{
				g[i].c='b';
				for(node tmp:g[i].adj)
					if(tmp.d==curr)
						{
						tmp.c='b';
						tmp.alive=false;
						for(node tmp1:tmp.adj)
							tmp1.d--;
						
						curr--;
						}
				
				g[i].alive=false;
				for(node tmp1:g[i].adj)
					tmp1.d--;
				break;
			}
		}
		int counter=2;
		for(int i=0;i<n;i++)
		{
			if(g[i].alive)
			{
				if(counter==0)
				{
					System.out.println("No");
					System.exit(0);
				}
				char a;
				if(counter==2)
					a='a';
				else
					a='c';
				dfs(g[i],a,g[i].d);
				counter--;
			}
		}
		System.out.println("Yes");
		for(int i=0;i<n;i++)
			System.out.print(g[i].c);
		
	}
	public static void dfs(node n,char c,int deg)
	{
		//System.out.println(n.n);
		if(!n.alive)
			return;
		if(n.d!=deg)
		{
			System.out.println("No");
			System.exit(0);
		}
		n.c=c;
		n.alive=false;
		
		if(n.adj.isEmpty())
			return;
		
		for(node tmp1:n.adj){
			tmp1.d--;
		}
		for(node n1:n.adj)
			dfs(n1,c,deg-1);
	}
	
	static class node{
		ArrayList<node> adj;
		int n;
		int d;
		char c;
		boolean alive;
		public node(int n)
		{
			alive=true;
			d=0;
			this.n=n;
			adj=new ArrayList<node>();
		}
	}
	

	public static long gcd(long a, long b) 
	{

		if(a==b)
			return a;
		if(a==0)
			return b;
		if(b==0)
			return a;

		if(a>b)
			return gcd(b,a%b);
		else
			return gcd(a,b%a);
	}

	public static long modPow( long a, long b, long mod )
    {
    	long res=1,pow=a;
    	while( b>0 )
    	{
    		if( (b&1)==1 )
    		{
    			res = (pow*res)%mod;
    		}
    		pow = (pow*pow)%mod;
    		b=b/2;
    	}
    	return res;
    }
	
	static class nck
	{
		public static long ncr[][];
		
		public nck(long z,long p)//z==maxvalue  p==mod
		{
			generateNCR(z, p);
		}
		
	    private static void generateNCR( long z, long p )
	    {
	    	ncr = new long[(int)z+1][(int)z+1];
	    	ncr[0][0] = 1;
	    	for(int i=1; i<=z; i++)
	    	{
	    		ncr[i][0] = 1;
	    		ncr[0][i] = 0;
	    	}
	        for(int i=1; i<=z; i++)
	        {
	            for(int j=1; j<=z; j++)
	            {
	                ncr[i][j] = (ncr[i-1][j] + ncr[i-1][j-1]) % p;
	            }
	        }
	    }

	    private static long lucas(long n, long k, long p) 
	    {
	    
	        long ans = 1;
	        
	        while(n>0)
	        {
	            int N = (int)(n%p);
	            int K = (int)(k%p);
	            if(K>N)
	                return 0;
	            ans *= (ncr[N][K] );
	            n /= p;
	            k /= p;
	        }
	        return ans % p;
		
    }
	}
	
	
	

	
	

	
	
	
	
	static class SegTree
	{
		static int seg[];
		static int dp[];
		int n;
		int d;
		public SegTree(int[] arr,int defaul)
		{
			this.dp=arr;
			n=dp.length;
			seg=new int[4*arr.length];
			build(1,0,n);
			this.d=defaul;
		}

		public void build(int id,int l,int r){
			if(r-l<2){
				seg[id]=dp[l];
				return;
			}
			int mid=(l+r)/2;
			build(2*id,l,mid);
			build(2*id+1,mid,r);
			seg[id]=merge(seg[2*id],seg[2*id+1]);
		}

		public int query(int l,int r)
		{
			return givemin(1,l,r,0,n);
		}

		public  int givemin(int id,int x,int y,int l,int r){
			if(x<=l && y>=r){
				return seg[id];
			}
			if(y<=l || r<=x){
				return d;
			}
			int mid=(l+r)/2;
			return merge(givemin(2*id, x, y, l, mid),givemin(2*id+1, x, y, mid, r));
		}

		public int upd(int id,int value)
		{
			return update(1,id,value,0,n);
		}
		public  int update(int id,int p,int value,int l,int r){
			if(r-l<2){
				dp[p]=value;
				seg[id]=value;
				return value;
			}
			int mid=(l+r)/2;
			if(p<mid){
				update(2*id, p, value, l, mid);
				seg[id]=merge(seg[2*id], seg[2*id+1]);
				return seg[id];
			}
			else{
				update(2*id+1, p, value, mid,r);
				seg[id]=merge(seg[2*id+1], seg[2*id]);
				return seg[id];

			}
		}

		public int merge(int l,int r)
		{

			//set initial value
			return l+r;
		}
	}


	static class UF {

		private int[] parent;  // parent[i] = parent of i
		private byte[] rank;   // rank[i] = rank of subtree rooted at i (never more than 31)
		private int count;   
		int[] seg;
		int max;
		// number of components
		public UF(int N) {
			if (N < 0) throw new IllegalArgumentException();
			count = N;
			parent = new int[N];
			rank = new byte[N];
			seg=new int[N];
			for (int i = 0; i < N; i++) {
				parent[i] = i;
				rank[i] = 0;
				seg[i]=1;
			}
			max=1;
		}

		public int find(int p) {
			if (p < 0 || p >= parent.length) throw new IndexOutOfBoundsException();
			while (p != parent[p]) {
				parent[p] = parent[parent[p]];    // path compression by halving
				p = parent[p];
			}
			return p;
		}
		public int count() {
			return count;
		}

		public boolean connected(int p, int q) {
			return find(p) == find(q);
		}

		public boolean union(int p, int q) {
			//    System.out.println("uf "+p+" "+q);
			int rootP = find(p);
			int rootQ = find(q);
			if (rootP == rootQ) return false;

			// make root of smaller rank point to root of larger rank
			if      (rank[rootP] < rank[rootQ])
			{
				parent[rootP] = rootQ;
				seg[rootQ]+=seg[rootP];
				max=Math.max(seg[rootQ],max);
			}
			else if (rank[rootP] > rank[rootQ])
			{
				parent[rootQ] = rootP;
				seg[rootP]+=seg[rootQ];
				max=Math.max(seg[rootP],max);
			}
			else {
				parent[rootQ] = rootP;
				rank[rootP]++;
				seg[rootP]+=seg[rootQ];
				max=Math.max(seg[rootP],max);
			}
			count--;
			return true;
		}

	}



	static class SparseTable
	{
		int[][] st;
		public SparseTable(int[] h){
			int n=h.length;
			int MAX_LOG=(int)Math.ceil(Math.log(n)/Math.log(2))+1;
			st=new int[n][MAX_LOG];
			for (int i = 0; i < n; i++)
				st[i][0] = h[i];
			for (int j = 1; (1 << j) <= n; j++)
			{
				for (int i = 0;( i + (1 << j) - 1 )< n; i++)
				{
					st[i][j]=Math.max(st[i][j-1],st[i+(1<<(j-1))][j-1]);
				}
			}
		}
		public int query(int l,int r)
		{
			int x=(int )Math.floor((Math.log(r-l+1)/Math.log(2)));
			int ans=Math.max(st[l][x],st[r-(1<<x)+1][x]);
			return ans;
		}

	}

	public static class FasterScanner {
		private byte[] buf = new byte[1024];
		private int curChar;
		private int numChars;

		public int read() {
			if (numChars == -1)
				throw new InputMismatchException();
			if (curChar >= numChars) {
				curChar = 0;
				try {
					numChars = System.in.read(buf);
				} catch (IOException e) {
					throw new InputMismatchException();
				}
				if (numChars <= 0)
					return -1;
			}
			return buf[curChar++];
		}

		public String nextLine() {
			int c = read();
			while (isSpaceChar(c))
				c = read();
			StringBuilder res = new StringBuilder();
			do {
				res.appendCodePoint(c);
				c = read();
			} while (!isEndOfLine(c));
			return res.toString();
		}

		public String nextString() {
			int c = read();
			while (isSpaceChar(c))
				c = read();
			StringBuilder res = new StringBuilder();
			do {
				res.appendCodePoint(c);
				c = read();
			} while (!isSpaceChar(c));
			return res.toString();
		}

		public long nextLong() {
			int c = read();
			while (isSpaceChar(c))
				c = read();
			int sgn = 1;
			if (c == '-') {
				sgn = -1;
				c = read();
			}
			long res = 0;
			do {
				if (c < '0' || c > '9')
					throw new InputMismatchException();
				res *= 10;
				res += c - '0';
				c = read();
			} while (!isSpaceChar(c));
			return res * sgn;
		}

		public int nextInt() {
			int c = read();
			while (isSpaceChar(c))
				c = read();
			int sgn = 1;
			if (c == '-') {
				sgn = -1;
				c = read();
			}
			int res = 0;
			do {
				if (c < '0' || c > '9')
					throw new InputMismatchException();
				res *= 10;
				res += c - '0';
				c = read();
			} while (!isSpaceChar(c));
			return res * sgn;
		}

		public int[] nextIntArray(int n) {
			int[] arr = new int[n];
			for (int i = 0; i < n; i++) {
				arr[i] = nextInt();
			}
			return arr;
		}

		public long[] nextLongArray(int n) {
			long[] arr = new long[n];
			for (int i = 0; i < n; i++) {
				arr[i] = nextLong();
			}
			return arr;
		}

		private boolean isSpaceChar(int c) {
			return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
		}

		private boolean isEndOfLine(int c) {
			return c == '\n' || c == '\r' || c == -1;
		}
	}

}
