code force 403C.C. Andryusha and Colored Balloons-优快云博客

解决Andryusha在公园中挂彩色气球的问题，确保相邻路径上的气球颜色互不相同，且使用尽可能少的不同颜色。通过DFS算法实现有效的颜色分配。

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 if a, b and c are distinct squares that a and b have a direct path between them, and b 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!

Input

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 integers x 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.

Output

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

In the second line print n integers, the i-th of them should be equal to the balloon color on the i-th square. Each of these numbers should be within range from 1 to k.

Examples

input

3
2 3
1 3

output

3
1 3 2

input

output

5
1 3 2 5 4

input

output

3
1 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.

$\text{[math]}$ 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. $\text{[math]}$ 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. $\text{[math]}$ Illustration for the third sample.

/*
题意：给你一棵数n个节点，n-1条边，让你进行染色，要求是a-b,b-c三点的颜色不同，让你给出配色方案，配色从1开始

初步思路：从1点开始搜，每个节点的染色和父亲，祖父的颜色不相同
*/

#include<bits/stdc++.h>
using namespace std ;
#define pb push_back
int col=1;//表示总染色的种类
int color[200010];
vector<int> edge[200010];
void dfs(int u,int fa){//父节点，祖父节点
    int colv=0;//将colv定义在循环外边，神来之笔，这样兄弟的颜色肯定是不会重复的了
    for(int i=0;i<edge[u].size();i++){//遍历儿子节点
        int v=edge[u][i];
        if(color[v]) continue;
        // cout<<v<<endl;
        colv++;
        while(colv==color[u]||colv==color[fa]) colv++;
        col=max(col,colv);
        color[v]=colv;
        dfs(v,u);
    }
}
int n,u,v;
int main(){
    // freopen("in.txt","r",stdin);
    scanf("%d",&n);
    for(int i=0;i<n-1;i++){
        scanf("%d%d",&u,&v);
        edge[u].pb(v);
        edge[v].pb(u);
    }//建图
    color[1]=1;
    dfs(1,1);
    printf("%d\n",col);
    for(int i=1;i<=n;i++){
        printf(i==1?"%d":" %d",color[i]);
    }
    printf("\n");
    return 0;
}