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.math.BigInteger;
import java.util.*;
import java.util.Map.Entry;

import static java.lang.Math.*;

public class Main {


	public static void solve(FastScanner fscan, BufferedWriter bw) throws IOException {
		final int n = fscan.nextInt(),
			e = fscan.nextInt();
		boolean graph[][] = new boolean[n][n];
		char answer[] = new char[n];
		int digree[] = new int[n];
		int aCount = 0, bCount=0, cCount = 0;

		
		for (int i = 0; i < e; i++) 
		{
			int s = fscan.nextInt() - 1;
			int d = fscan.nextInt() - 1;
			
			graph[s][d] = graph[d][s] = true;
			digree[s]++;
			digree[d]++;
		}
		
		for (int i = 0; i < n; i++) {
			if(digree[i] == n-1)
			{
				answer[i] = 'b';
				bCount++;
			}
		}
		
		int aEdge = -1;
		
		for(int i = 0 ; i< n ; i++)
		{
			if(answer[i] == 0) {
				answer[i] = 'a';
				aCount++;
				aEdge = i;
				break;
			}
		}
		if(aEdge != -1) {
			for (int k = 0; k < n; k++) {
				if(graph[aEdge][k] && answer[k] == 0)
				{
					answer[k] = 'a';
					aCount++;
				}
			}
		}

		
		int cEdge = -1;		
		for(int i = 0 ; i< n ; i++)
		{
			if(answer[i] == 0) {
				answer[i] = 'c';
				cCount++;
				cEdge = i;
				break;
			}
		}
		
		if(cEdge != -1) {
			for (int k = 0; k < n; k++) {
				if(graph[cEdge][k] && answer[k] == 0)
				{
					answer[k] = 'c';
					cCount++;							
				}
			}
		}

		boolean isValidGraph = aCount + bCount + cCount == n;
		validate:for (int i = 0; i < n; i++) {
			switch(answer[i])
			{
			case 'a':
				if(digree[i] != aCount + bCount -1){
					isValidGraph = false;
					break validate;
				}
				break;
			case 'b':
				if(digree[i] != n-1){
					isValidGraph = false;
					break validate;
				}
				break;
			case 'c':
				if(digree[i] != bCount + cCount -1){
					isValidGraph = false;
					break validate;
				}
				break;
			}
		}
		
		if(!isValidGraph)
		{
			bw.write("No");
		}
		else
		{
			bw.write("Yes\n");
			for (int i = 0; i < n; i++) {
				bw.append(answer[i]);
			}
		}
	}

	/**************** ENTER HERE   ********************/
	public static void main(String[] args) throws Exception {
		
		if(args.length != 0){
			test();
			return;
		}
		
		FastScanner fscan = new FastScanner(System.in);
		BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
		solve(fscan,bw);
		bw.close();
	}
	
	/****   Test Static test cases      **/
	public static void test() throws IOException{
		String testcase[] = new String[]{};
		String answer[] = new String[]{};
		
		for (int i = 0; i < testcase.length; i++) {
			ByteArrayOutputStream result = new ByteArrayOutputStream();
			FastScanner fscan = new FastScanner(new ByteArrayInputStream(testcase[i].getBytes()));
			BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(result));
			
			solve(fscan, bw);
			
			String r = new String(result.toByteArray());
			if(! r.equals(answer[i])){
				String q = testcase[i];
				String a = answer[i];

				System.err.println("Wrong Answer on test:"+(i+1));
				System.err.println("Q:"+q);
				System.err.println("A:"+a);
				System.err.println("R:"+r);
			}
			
		}

	}
}






class IO {
	public static void readIntArray(int a[], int N, FastScanner fscan)
			throws IOException {
		readIntArray(a, N, 0, fscan);
	}

	public static void readIntArray(int a[], int N, int start, FastScanner fscan)
			throws IOException {
		for (int i = 0; i < N; i++) {
			a[start + i] = fscan.nextInt();
		}
	}

	public static void readArray(long a[], int N, FastScanner fscan)
			throws IOException {
		readArray(a, N, 0, fscan);
	}

	public static void readArray(long a[], int N, int start, FastScanner fscan)
			throws IOException {
		for (int i = 0; i < N; i++) {
			a[start + i] = fscan.nextInt();
		}
	}
}

class FastScanner {
	InputStream is;
	byte buff[] = new byte[1024];
	int currentChar = -1;
	int buffChars = 0;

	public FastScanner(InputStream inputStream) {
		is = inputStream;
	}

	public boolean hasNext() throws IOException {
		return currentChar == -1 || buffChars > 0;
	}

	public char nextChar() throws IOException {
		// if we already have that next char read, just return else input
		if (currentChar == -1 || currentChar >= buffChars) {
			currentChar = 0;
			buffChars = is.read(buff);
		}

		if (buffChars <= 0) {
			throw new RuntimeException("No char found...");
		}

		int ch;
		while (isSpace(ch = nextCharAsInt()))
			;
		return (char) ch;
	}

	public int nextCharAsInt() throws IOException {
		// if we already have that next char read, just return else input
		if (currentChar == -1 || currentChar >= buffChars) {
			currentChar = 0;
			buffChars = is.read(buff);
		}

		if (buffChars <= 0) {
			return -1;
		}
		return (char) buff[currentChar++];
	}

	public String nextLine() throws IOException {
		StringBuilder bldr = new StringBuilder();
		int ch;
		while (isLineEnd(ch = nextCharAsInt()))
			// ignore empty lines ???
			;

		do {
			bldr.append((char) ch);
		} while (!isLineEnd(ch = nextCharAsInt()));

		return bldr.toString();
	}

	public String nextString() throws IOException {
		StringBuilder bldr = new StringBuilder();
		int ch;
		while (isSpace(ch = nextCharAsInt()))
			;

		do {
			bldr.append((char) ch);
		} while (!isSpace(ch = nextCharAsInt()));

		return bldr.toString();
	}

	public int nextInt() throws IOException {
		// considering ASCII files--> 8 bit chars, unicode files has 16 bit
		// chars (byte1 then byte2)
		int result = 0;
		int sign = 1;
		int ch;
		while (isSpace(ch = nextCharAsInt()))
			;

		if (ch == '-') {
			sign = -1;
			ch = nextCharAsInt();
		}

		do {
			if (ch < '0' || ch > '9')
				throw new NumberFormatException("Found '" + ch
						+ "' while parsing for int.");

			result *= 10;
			result += ch - '0';
		} while (!isSpace(ch = nextCharAsInt()));

		return sign * result;
	}

	public long nextLong() throws IOException {
		// considering ASCII files--> 8 bit chars, unicode files has 16 bit
		// chars (byte1 then byte2)
		long result = 0;
		int sign = 1;
		int ch;
		while (isSpace(ch = nextCharAsInt()))
			;

		if (ch == '-') {
			sign = -1;
			ch = nextCharAsInt();
		}

		do {
			if (ch < '0' || ch > '9')
				throw new NumberFormatException("Found '" + ch
						+ "' while parsing for int.");

			result *= 10;
			result += ch - '0';
		} while (!isSpace(ch = nextCharAsInt()));

		return sign * result;
	}

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

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

	public void close() throws IOException {
		is.close();
	}
}
