CodeForces - 624C Graph and String

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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 $\text{[math]}$ — 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.

题意：给定一个无向图，每个顶点对应abc中的一个字符，其中a必须和a,c相连，b必须和a,b,c相连，c必须和b,c相连，问能否找到每个顶点对应字符满足上述关系

思路：突破口在发现b必须和其他所有的字符都相连，也就是每个b字符的度数都为n-1，剩下的要么是a要么是c，而a和c又不相连，所以找一个没被染色的顶点，令其是a，除b外，与它相连的都是a，不相连的都是c，最后在遍历一遍看是不是所有的点都符合题目中的关系。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <queue>
#include <stack>
#include <map>
#include <cmath>
#include <vector>
#define max_ 600
#define inf 0x3f3f3f3f
#define ll long long
using namespace std;

int n,m;
int mp[max_][max_];
int col[max_];
int deg[max_];
int main(int argc, char const *argv[]) {
    cin>>n>>m;
    while(m--)
    {
        int x,y;
        cin>>x>>y;
        deg[x]++;
        deg[y]++;
        mp[x][y]=mp[y][x]=1;
    }
    for(int i=1;i<=n;i++)
    {
        mp[i][i]=1;
        if(deg[i]==n-1)
        col[i]=2;
    }
    for(int i=1;i<=n;i++)
    {
        if(col[i]==0)
        {
            col[i]=1;
            for(int j=i+1;j<=n;j++)
            {
                if(col[j]==0)
                {
                    if(mp[i][j])
                    {
                        col[j]=1;
                    }
                    else
                    {
                        col[j]=3;
                    }
                }
            }
            break;
        }
    }
    int f=0;
    for(int i=1;i<=n;i++)
    {
        if(col[i]==1)
        {
            for(int j=i;j<=n;j++)
            {
                if((col[j]==3&&mp[i][j])||(col[j]==1&&mp[i][j]==0))
                {
                    f=1;
                    break;
                }
            }
            if(f)
            break;
        }
        else if(col[i]==3)
        {
            for(int j=i;j<=n;j++)
            {
                if((col[j]==1&&mp[i][j])||(col[j]==3&&mp[i][j]==0))
                {
                    f=1;
                    break;
                }
            }
            if(f)
            break;
        }
    }
    if(f)
    {
        printf("No\n" );
    }
    else
    {
        printf("Yes\n" );
        for(int i=1;i<=n;i++)
        {
            if(col[i]==1)
            printf("a" );
            else if(col[i]==2)
            printf("b" );
            else if(col[i]==3)
            printf("c" );
        }
        printf("\n" );
    }
    return 0;
}