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.BufferedInputStream;
import java.io.IOException;
import java.io.PrintWriter;
import java.text.DecimalFormat;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Comparator;
import java.util.HashMap;
import java.util.HashSet;
import java.util.InputMismatchException;
import java.util.LinkedList;
import java.util.Map.Entry;
import java.util.Queue;
import java.util.Random;
import java.util.Set;
import java.util.SortedSet;
import java.util.Stack;
import java.util.TreeMap;
import java.util.TreeSet;

public class Main
{	
	/********************************************** a list of common variables **********************************************/
	private MyScanner scan = new MyScanner();
	private PrintWriter out = new PrintWriter(System.out);
	private final double PI = Math.acos(-1.0);
	private final int SIZEN = (int)(1e5);
	private final int MOD = (int)(1e9 + 7);
	private final int[] DX = {0, 1, 0, -1}, DY = {-1, 0, 1, 0};
	private ArrayList<Integer>[] edge;

	public void foo()
	{
		int n = scan.nextInt();
		int m = scan.nextInt();
		boolean[][] edge = new boolean[n][n];
		for(int i = 0;i < m;++i)
		{
			int u = scan.nextInt() - 1;
			int v = scan.nextInt() - 1;
			edge[u][v] = true;
			edge[v][u] = true;
		}
		char[] ans = new char[n];
		for(int i = 0;i < n;++i)
		{
			boolean hasNot = false;
			for(int j = 0;j < n;++j)
			{
				if(i != j && !edge[i][j])
				{
					hasNot = true;
					break;
				}
			}
			if(!hasNot)
			{
				continue;
			}
			Queue<Integer> q = new LinkedList<Integer>();
			q.offer(i);
			while(!q.isEmpty())
			{
				int j = q.poll();
				if(0 == ans[j])
				{
					for(int k = 0;k < n;++k)
					{
						if(k != j && !edge[j][k] && ans[k] != 0)
						{
							ans[j] = 'a' == ans[k] ? 'c' : 'a';
							break;
						}
					}
					if(0 == ans[j])
					{
						ans[j] = 'a';
					}
				}
				for(int k = 0;k < n;++k)
				{
					if(k != j && !edge[j][k])
					{
						if(0 == ans[k])
						{
							ans[k] = 'a' == ans[j] ? 'c' : 'a';
							q.offer(k);
						}
						else if(ans[k] == ans[j])
						{
							out.println("NO");
							return;
						}
					}		
				}
			}
		}
		for(int i = 0;i < n;++i)
		{
			for(int j = i + 1;j < n;++j)
			{
				if(edge[i][j] && ans[i] != 0 && ans[j] != 0 && 2 == Math.abs(ans[i] - ans[j]))
				{
					out.println("NO");
					return;
				}
			}
		}
		for(int i = 0;i < n;++i)
		{
			if(0 == ans[i])
			{
				ans[i] = 'b';
			}
		}
		out.println("YES");
		out.println(ans);
	} 

	public static void main(String[] args)
	{
		Main m = new Main();
		m.foo();
		m.out.close();
	}

	/********************************************** a list of common algorithms **********************************************/
	/**
	 * 1---Get greatest common divisor
	 * @param a :	first number
	 * @param b :	second number
	 * @return		greatest common divisor
	 */
	public long gcd(long a, long b)
	{
		return 0 == b ? a : gcd(b, a % b);
	}
	
	/**
	 * 2---Get the distance from a point to a line
	 * @param x1	the x coordinate of one endpoint of the line
	 * @param y1	the y coordinate of one endpoint of the line
	 * @param x2	the x coordinate of the other endpoint of the line
	 * @param y2	the y coordinate of the other endpoint of the line
	 * @param x		the x coordinate of the point 
	 * @param y		the x coordinate of the point
	 * @return		the distance from a point to a line
	 */
	public double getDist(long x1, long y1, long x2, long y2, long x, long y)
	{
		long a = y2 - y1;
		long b = x1 - x2;
		long c = y1 * (x2 - x1) - x1 * (y2 - y1);
		return Math.abs(a * x + b * y + c) / Math.sqrt(a * a + b * b);
	}
	
	/**
	 * 3---Get the distance from one point to a segment (not a line)
	 * @param x1	the x coordinate of one endpoint of the segment
	 * @param y1	the y coordinate of one endpoint of the segment
	 * @param x2	the x coordinate of the other endpoint of the segment
	 * @param y2	the y coordinate of the other endpoint of the segment
	 * @param x		the x coordinate of the point
	 * @param y		the y coordinate of the point
	 * @return		the distance from one point to a segment (not a line)
	 */
	public double ptToSeg(long x1, long y1, long x2, long y2, long x, long y)
	{
		double cross = (x2 - x1) * (x - x1) + (y2 - y1) * (y - y1);
		if(cross <= 0)
		{
			return (x - x1) * (x - x1) + (y - y1) * (y - y1);
		}
		double d = (x2 - x1) * (x2 - x1) + (y2 - y1) * (y2 - y1);
		if(cross >= d) 
		{
			return (x - x2) * (x - x2) + (y - y2) * (y - y2);
		}
		double r = cross / d;
		double px = x1 + (x2 - x1) * r;
		double py = y1 + (y2 - y1) * r;
		return (x - px) * (x - px) + (y - py) * (y - py);
	}
	
	/**
	 * 4---KMP match, i.e. kmpMatch("abcd", "bcd") = 1, kmpMatch("abcd", "bfcd") = -1.
	 * @param t:	String to match.
	 * @param p:	String to be matched.
	 * @return		if can match, first index; otherwise -1.
	 */
	public int kmpMatch(char[] t, char[] p)
	{
		int n = t.length;
		int m = p.length;
		int[] next = new int[m + 1];
		next[0] = -1;
		int j = -1;
		for(int i = 1;i < m;++i)
		{
			while(j >= 0 && p[i] != p[j + 1])
			{
				j = next[j];
			}
			if(p[i] == p[j + 1])
			{
				++j;
			}
			next[i] = j;
		}
		j = -1;
		for(int i = 0;i < n;++i)
		{
			while(j >= 0 && t[i] != p[j + 1])
			{
				j = next[j];
			}
			if(t[i] == p[j + 1])
			{
				++j;
			}
			if(j == m - 1)
			{
				return i - m + 1;
			}
		}
		return -1;
	}
	
	class MyScanner
	{
	    private byte[] buf = new byte[1024];
	    private int curChar;
	    private int numChars;
	    BufferedInputStream bis = new BufferedInputStream(System.in);

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

	    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 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 double nextDouble() 
	    {
	        int c = read();
	        while (isSpaceChar(c))
	        {
	        	c = read();
	        }
	        int sgn = 1;
	        if (c == '-')
	        {
	            sgn = -1;
	            c = read();
	        }
	        double res = 0;
	        while (!isSpaceChar(c) && c != '.') 
	        {
	            if (c == 'e' || c == 'E')
	            {
	            	return res * Math.pow(10, nextInt());
	            }
	            if (c < '0' || c > '9')
	            {
	            	throw new InputMismatchException();
	            }
	            res *= 10;
	            res += c & 15;
	            c = read();
	        }
	        if (c == '.') 
	        {
	            c = read();
	            double m = 1;
	            while (!isSpaceChar(c)) 
	            {
	                if (c == 'e' || c == 'E')
	                {
	                	return res * Math.pow(10, nextInt());
	                }
	                if (c < '0' || c > '9')
	                {
	                	throw new InputMismatchException();
	                }
	                m /= 10;
	                res += (c & 15) * m;
	                c = read();
	            }
	        }
	        return res * sgn;
	    }
	    
	    public String next()
	    {
	        int c = read();
	        while (isSpaceChar(c))
	        {
	        	c = read();
	        }
	        StringBuilder res = new StringBuilder();
	        do 
	        {
	            res.appendCodePoint(c);
	            c = read();
	        } while (!isSpaceChar(c));
	        return res.toString();
	    }

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