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.
The first line of the input contains two integers n and
m 
— 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.
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.
2 1 1 2
Yes aa
4 3 1 2 1 3 1 4
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.
#include<bits/stdc++.h>
using namespace std;
int n,m;
int color[5010];//染色的顏色，-1爲初始值，0 1爲染色的顏色
int a[5010][5010];//鄰接矩陣存圖
vector <int> G[5010];//vector模擬鄰接表，存儲補圖
bool is_ac[5010];//判斷這個點是否爲B
bool colored[5010];//代表這個點是否拓展過了（是否從這個點出發前往它的相鄰點了）
bool ans=true;//初始化可以成功
void paint(int i,int x)//將節點i染成x顏色
{
if((color[i]!=-1)&&(color[i]!=x))//如果i已經有了顏色，而且不是x
{
ans=false;//肯定失敗
}
color[i]=x;
if(colored[i]==true)
return ;//如果已經拓展過不用重複拓展
colored[i]=true;//設置已經拓展的tag
for(auto j:G[i])
paint(j,1-x);//dfs
}
int main(void)
{
int u,v;
cin>>n>>m;
memset(color,-1,sizeof(color));//顏色全部初始化
memset(a,-1,sizeof(a));
memset(is_ac,0,sizeof(is_ac));//都初始化爲是B
memset(colored,0,sizeof(colored));
for(int i=0;i<m;i++)
{
scanf("%d%d",&u,&v);
a[u][v]=a[v][u]=1;
}
for(int i=1;i<=n;i++)
{
for(int j=1;j<=n;j++)
{
if(i==j)
continue;
if(a[i][j]==-1)//求補圖
{
G[i].push_back(j);
is_ac[i]=is_ac[j]=true;//這些點不會是B，是a或者C
}
}
}
for(int i=1;i<=n;i++)//對所有連通域進行染色
{
if(is_ac[i]==false)//是B的話直接不用考慮
continue;
int temp_color=color[i];
if(temp_color==-1)//要是這個點的顏色還是初始的-1的話
temp_color=0;
if(colored[i]==false)//如果這個點還沒有拓展過/處於不同的連通域
paint(i,temp_color);
}
if(ans)//如果染色完畢，答案爲true
{
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
{
if((a[i][j]==1)&&(color[i]+color[j]==1))//對第一個條件檢查
ans=false;
}
}
if(ans)
cout<<"Yes"<<endl;
else
cout<<"No"<<endl;
if(ans)
{
for(int i=1;i<=n;i++)
{
if(is_ac[i]==false)
cout<<"b";
else if(color[i])
cout<<"a";
else
cout<<"c";
}
}
return 0;
}
