Codeforces Round #403 (Div. 2) C. Andryusha and Colored Balloons

C. Andryusha and Colored Balloons

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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

5
2 3
5 3
4 3
1 3

output

5
1 3 2 5 4 

input

5
2 1
3 2
4 3
5 4

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.

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.

题意：给你一棵树，树上相邻的三个点不能涂同样的颜色，问你最少需要多少种颜色并把其中一种涂法写出来。

思路：明显要树形dp，不过此dp排除相邻3个点的颜色的方法略强，表示我不看题解想不出。给代码自己体会。

#include<iostream>  
#include<cmath>  
#include<queue>  
#include<cstdio>  
#include<queue>  
#include<algorithm>  
#include<cstring>  
#include<string>  
#include<utility>
#include<map>
#define maxn 200005  
#define inf 0x3f3f3f3f  
using namespace std; 
typedef long long LL; 
const double eps=1e-8;
struct node{
    int v,next;    
}p[maxn<<1];
int head[maxn],ans[maxn],length;
void dfs(int x,int fa){
    int number=1;
    for(int i=head[x];~i;i=p[i].next){
	if(p[i].v==fa)
	    continue;
	while(number==ans[x]||number==ans[fa])
	    number++;
	ans[p[i].v]=number++;
	length=max(length,number-1);
	dfs(p[i].v,x);
    }   
}
int main(){
    memset(head,-1,sizeof(head));
    int n;
    scanf("%d",&n);
    int len=0;
    for(int i=0;i<n-1;i++){
	int u,v;
	scanf("%d%d",&u,&v);
	p[len].v=v;
	p[len].next=head[u];
	head[u]=len++;	
	p[len].v=u;
	p[len].next=head[v];
	head[v]=len++;
    }
    ans[1]=1;
    length=1;
    dfs(1,0);
    printf("%d\n%d",length,ans[1]);
    for(int i=2;i<=n;i++){
	printf(" %d",ans[i]);	
    }
    printf("\n");
}