Problem:
Score : 600 points

Problem Statement

We have a tree with N vertices. The vertices are numbered 1 to N, and the i-th edge connects Vertex a_i and Vertex b_i.

Takahashi loves the number 3. He is seeking a permutation p_1, p_2, \ldots , p_N of integers from 1 to N satisfying the following condition:

For every pair of vertices (i, j), if the distance between Vertex i and Vertex j is 3, the sum or product of p_i and p_j is a multiple of 3.

Here the distance between Vertex i and Vertex j is the number of edges contained in the shortest path from Vertex i to Vertex j.

Help Takahashi by finding a permutation that satisfies the condition.

Constraints

2\leq N\leq 2\times 10^5

1\leq a_i, b_i \leq N

The given graph is a tree.

Input

Input is given from Standard Input in the following format:

N
a_1 b_1
a_2 b_2
\vdots
a_{N-1} b_{N-1}

Output

If no permutation satisfies the condition, print -1.

Otherwise, print a permutation satisfying the condition, with space in between.
If there are multiple solutions, you can print any of them.

Sample Input 1

5
1 2
1 3
3 4
3 5

Sample Output 1

1 2 5 4 3

The distance between two vertices is 3 for the two pairs (2, 4) and (2, 5).

p_2 + p_4 = 6

p_2\times p_5 = 6

Thus, this permutation satisfies the condition.

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <limits.h>
#include <stdint.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <unistd.h>
#include <math.h>

#define P 1000000007

int comp(const void *a, const void *b){return *(int*)a-*(int*)b;}

int e[400000][2];
int d[200001];
int search(int x, int l, int r){
	while(l<=r){
		int m=(l+r)/2;
		if(e[m][0]>=x) r=m-1; else l=m+1;
	}
	return l;
}

int main(void){
	int n, i;
	scanf("%d", &n);
	for(i=0; i<n-1; i++){
		int A, B;
		scanf("%d%d", &A, &B);
		e[i*2][0]=A; e[i*2][1]=B;
		e[i*2+1][0]=B; e[i*2+1][1]=A;
	}
	qsort(e, (n-1)*2, sizeof(int)*2, comp);

	int s[200000], S=0;
	d[1]=0;
	s[S++]=1;
	while(S>0){
		int p=s[--S];
		i=search(p, 0, (n-1)*2-1);
		for(; e[i][0]==p; i++){
			if(e[i][1]!=1 && d[e[i][1]]==0){
				d[e[i][1]] = d[p] + 1;
				s[S++]=e[i][1];
			}
		}
	}

	int b=0, c=0;
	for(i=1; i<=n; i++){
		if(d[i]%2==0){
			b++;
		}else{
			c++;
		}
	}
	if(b<c){i=b; b=c; c=i;}
	if(b>n/3+(n+2)/3){
		int a=0;
		b=(n+2)/3; c=(n+1)/3;
		for(i=1; i<=n; i++){
			if(d[i]%2==0){
				if(b>0){
					b--;
					printf("%d ", b*3+1);
				}else if(c>0){
					c--;
					printf("%d ", c*3+2);
				}else{
					a++;
					printf("%d", a*3);
				}
			}else{
				a++;
				printf("%d", a*3);
			}
		}
	}else if(c>n/3+(n+1)/3){
		int a=0;
		b=(n+2)/3; c=(n+1)/3;
		for(i=1; i<=n; i++){
			if(d[i]%2==0){
				a++;
				printf("%d", a*3);
			}else{
				if(b>0){
					b--;
					printf("%d ", b*3+1);
				}else if(c>0){
					c--;
					printf("%d ", c*3+2);
				}else{
					a++;
					printf("%d", a*3);
				}
			}
		}
	}else{
		int a=0;
		b=(n+2)/3; c=(n+1)/3;
		for(i=1; i<=n; i++){
			if(d[i]%2==0){
				if(b>0){
					b--;
					printf("%d ", b*3+1);
				}else{
					a++;
					printf("%d", a*3);
				}
			}else{
				if(c>0){
					c--;
					printf("%d ", c*3+2);
				}else{
					a++;
					printf("%d", a*3);
				}
			}
		}
	}
	putchar(10);
}