Andryusha goes through a park each day. The squares and paths between them look boring to Andryusha, so he decided to decorate them.

The park consists of n squares connected with(n?-?1) bidirectional paths in such a way that any square is reachable from any other using these paths. Andryusha decided to hang a colored balloon at each of the squares. The baloons' colors are described by positive integers, starting from 1. In order to make the park varicolored, Andryusha wants to choose the colors in a special way. More PRecisely, he wants to use such colors that ifa, b andc are distinct squares that a and b have a direct path between them, andb and c have a direct path between them, then balloon colors on these three squares are distinct.

Andryusha wants to use as little different colors as possible. Help him to choose the colors!

The first line contains single integer n (3?≤?n?≤?2·10⁵) — the number of squares in the park.

Each of the next (n?-?1) lines contains two integersx and y (1?≤?x,?y?≤?n) — the indices of two squares directly connected by a path.

It is guaranteed that any square is reachable from any other using the paths.

In the first line print single integer k — the minimum number of colors Andryusha has to use.

In the second line print n integers, thei-th of them should be equal to the balloon color on thei-th square. Each of these numbers should be within range from1 to k.

32 31 3Output
31 3 2 Input
52 35 34 31 3Output
51 3 2 5 4 Input
52 13 24 35 4Output
31 2 3 1 2 Note
In the first sample the park consists of three squares: 1?→?3?→?2. Thus, the balloon colors have to be distinct.
Illustration for the first sample. 
In the second example there are following triples of consequently connected squares:
1?→?3?→?2 1?→?3?→?4 1?→?3?→?5 2?→?3?→?4 2?→?3?→?5 4?→?3?→?5 We can see that each pair of squares is encountered in some triple, so all colors have to be distinct.Illustration for the second sample. 
In the third example there are following triples: 
1?→?2?→?3 2?→?3?→?4 3?→?4?→?5 We can see that one or two colors is not enough, but there is an answer that uses three colors only.Illustration for the third sample.
題目大意：
給你N個(gè)點(diǎn)，有N-1條邊，構(gòu)成一棵樹，要求任意一條長度為2的鏈（包含三個(gè)點(diǎn)）中的三個(gè)點(diǎn)，顏色都不能相同。
問最少需要幾種顏色就足夠了，并且要求輸出一種可行解。
思路：
顯然，max（degree【i】）+1就是需要的最少顏色個(gè)數(shù)。
那么接下來我們只要隨便Dfs染染色就行了。
#include<stdio.h>#include<string.h>#include<vector>using namespace std;vector<int >mp[300050];int color[300050];int degree[300050];int ans;void Dfs(int u,int from){    int tmp=1;    for(int i=0;i<mp[u].size();i++)    {        int v=mp[u][i];        if(v==from)continue;        else        {            for(int j=tmp;j<=ans;j++)            {                if(j!=color[u])                {                    if(from==-1)                    {                        color[v]=j;                        tmp=j+1;                        break;                    }                    else if(j!=color[from])                    {                        color[v]=j;                        tmp=j+1;                        break;                    }                }            }            Dfs(v,u);        }    }}int main(){    int n;    while(~scanf("%d",&n))    {        for(int i=1;i<=n;i++)mp[i].clear();        memset(degree,0,sizeof(degree));        for(int i=0;i<n-1;i++)        {            int x,y;            scanf("%d%d",&x,&y);            mp[x].push_back(y);            mp[y].push_back(x);            degree[x]++;            degree[y]++;        }        ans=0;        for(int i=1;i<=n;i++)ans=max(ans,degree[i]);        ans++;        printf("%d/n",ans);        color[1]=1;        Dfs(1,-1);        for(int i=1;i<=n;i++)        {            printf("%d ",color[i]);        }        printf("/n");    }}