C. Andryusha and Colored Balloons--bfs

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.

题目链接：http://codeforces.com/contest/782/problem/C

题目的意思是让你染色，相邻的三个不可有相同的颜色，那么我们考虑记录每个节点的父节点，给这个节点的子节点染色，用vector排一下序即可。

我手贱排了一下序，因为他说每个节点要用它自己编号的颜色，我就手贱了，但是为什么不用排序呢？

代码：

#include <cstdio>#include <cstring>#include <iostream>#include <queue>#include <algorithm>#include <vector>using namespace std;const int maxn=500000;vector<int>G[maxn];struct node1{    int color;    int p1;    int p2;}xin[maxn];void Init(){    for(int i=0;i<=400000;i++){        xin[i].color=-1;        xin[i].p1=0;        xin[i].p2=0;    }}int mix;void bfs(int s){    xin[s].color=1;    queue<int>Q;    Q.push(s);    while(!Q.empty()){        int t=Q.front();        Q.pop();        int kk=1;        for(int i=0;i<(int)G[t].size();i++){            int v=G[t][i];            if(xin[v].color!=-1)                continue;            xin[v].p1=t;            xin[v].p2=xin[t].p1;            while(kk){                if(xin[t].color==kk||xin[xin[t].p1].color==kk){                    kk++;                    mix=max(kk,mix);                }                else{                    mix=max(kk,mix);                    xin[v].color=kk++;                    break;                }            }            Q.push(v);        }    }}int main(){    int n;    mix=0;    Init();    scanf("%d",&n);    for(int i=0;i<n-1;i++){        int x,y;        scanf("%d%d",&x,&y);        G[x].push_back(y);        G[y].push_back(x);    }    //for(int i=1;i<=n;i++){//就在这，我手贱的拍了一下序，注释掉这三行就对了    //    sort(G[i].begin(),G[i].end());   // }    bfs(1);    //for(int i=1;i<=n;i++){    //    printf("%d %d %d\n",i,xin[i].p1,xin[i].p2);    //}    cout<<mix<<endl;    for(int i=1;i<=n;i++){        printf(i==1?"%d":" %d",xin[i].color);    }    cout<<endl;    return 0;}