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

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

32 31 3

output

31 3 2 

input

52 35 34 31 3

output

51 3 2 5 4 

input

52 13 24 35 4

output

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.

由于只有n-1条边，且图连通，所以一定是一个树

dfs遍历树，形参记录前一个节点和当前节点

对每个点，搜索他连接的且没走的点，因为是树，所以时间很低

对于染色，只要保证每个点和他的父亲，父亲的父亲以及兄弟不同色即可，并且对于一棵树，树上的任何一点都可以做根节点，所以从哪个点开始搜都可以

对于有多少种颜色，可以求完染色数组，统计标记一下求得，也可以记录每个点的度，找到最大再加一，因为只要保证对于每个点所有连的边以及他本身这个点所有颜色都各不相同即可，也就是最大度数加上自己

#include<stdio.h>#include<string.h>#include<math.h>#include<string>#include<algorithm>#include<queue>#include<vector>#include<map>#include<set>#define eps 1e-9#define PI 3.141592653589793#define bs 1000000007#define bsize 256#define MEM(a) memset(a,0,sizeof(a))typedef long long ll;using namespace std;vector<int>w[200005];//存图int ans[200005];//染色数组void dfs(int zi,int node){int i,j=1;for(i=0;i<w[node].size();i++)if(!ans[w[node][i]]){while(j==ans[zi]||j==ans[node])j++;ans[w[node][i]]=j++;//j++作为下一个兄弟的染色初值dfs(node,w[node][i]);}return;}int main(){int n,x,y,i;int book[200005];MEM(book);scanf("%d",&n);for(i=0;i<n-1;i++){scanf("%d %d",&x,&y);w[x].push_back(y);w[y].push_back(x);book[x]++;book[y]++;}int maxx=0;for(i=1;i<=n;i++)maxx=max(maxx,book[i]);ans[1]=1;dfs(0,1);printf("%d\n",maxx+1);for(i=1;i<=n;i++)printf("%d ",ans[i]);return 0; }