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

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

题意：给出一个n个点n-1条边的树，要求距离小于等于2的点不能同色。问最少需要几种颜色以及任意一种方案。

题解：需要颜色的数量=最大度数+1，然后dfs染色就好。

代码：

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int N=300050;
vector<int >mp
;
int clr
;
int dge
;
int ans,n;
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!=clr[u])
{

if(from==-1)
{
clr[v]=j;
tmp=j+1;
break;
}
else  if(j!=clr[from])
{
clr[v]=j;
tmp=j+1;
break;
}
}
}
Dfs(v,u);
}
}
}
void input()
{
cin>>n;
for(int i=1; i<=n; i++)mp[i].clear();
memset(dge,0,sizeof(dge));
for(int i=0; i<n-1; i++)
{
int x,y;
cin>>x>>y;
mp[x].push_back(y);
mp[y].push_back(x);
dge[x]++;
dge[y]++;
}

}
void output()
{

ans=0;
for(int i=1; i<=n; i++)ans=max(ans,dge[i]);
ans++;
cout<<ans<<endl;
clr[1]=1;
Dfs(1,-1);
for(int i=1; i<=n; i++)
{
cout<<clr[i]<<" ";
}
}
int main()
{
input();
output();
}