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 <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#define ll long long
#define rep(i,l,r)for(ll i=(l);i<(r);i++)
#define repp(i,l,r,k)for(ll i=(l);i<(r);i+=(k))
#define INF ((1LL<<62)-(1LL<<31))
#define max(p,q)((p)>(q)?(p):(q))
#define min(p,q)((p)<(q)?(p):(q))
#define bit(n,m)(((n)>>(m))&1)
int upll(const void*a, const void*b){return*(ll*)a<*(ll*)b?-1:*(ll*)a>*(ll*)b?1:0;}
int downll(const void*a, const void*b){return*(ll*)a<*(ll*)b?1:*(ll*)a>*(ll*)b?-1:0;}
void sortup(ll*a,int n){qsort(a,n,sizeof(ll),upll);}
void sortdown(ll*a,int n){qsort(a,n,sizeof(ll),downll);}
ll pom(ll a,ll n,int m){ll x=1;for(a%=m;n;n/=2)n&1?x=x*a%m:0,a=a*a%m;return x;}
//#define MOD 998244353
#define MOD 1000000007
#define invp(a,p)pom(a,p-2,p)


//辺の情報を個別に持つタイプ
typedef struct edge{ll s,g,c;}E;
typedef struct graph{
	int vcnt,ecnt;
	E  e[400010];//適宜変える(ecnt)
	int id[200010];//適宜変える(vcnt)
}G;
G g;

int esort(const void*a,const void*b){
	E*p=(E*)a,*q=(E*)b;
	if((*p).s<(*q).s)return -1;
	if((*p).s>(*q).s)return  1;
	if((*p).g<(*q).g)return -1;
	return 1;
}
void readgraph(){
	//適宜変える
	ll n;
	scanf("%lld",&n);
	rep(i,0,n-1){
		ll x,y;
		scanf("%lld%lld",&x,&y);
		x--,y--;
		g.e[2*i].s=x;
		g.e[2*i].g=y;
		g.e[2*i+1].s=y;
		g.e[2*i+1].g=x;
	}
	g.vcnt=n;
	g.ecnt=2*n-2;
	qsort(g.e,g.ecnt,sizeof(E),esort);

	int p=0;
	rep(i,0,g.vcnt){
		while(p<g.ecnt&&g.e[p].s<i)p++;
		g.id[i]=p;
	}
	g.id[g.vcnt]=g.ecnt;//番兵
}


ll dp[200010];

ll dfs(int v,int pre,int c){
    printf("%d %d\n",v,c);
    dp[v]=c;
	rep(i,g.id[v],g.id[v+1]){
		if(g.e[i].g!=pre){
			dfs(g.e[i].g,v,c^1);
		}
	}
}

int main(){
    readgraph();
    dfs(0,-1,0);
    
    
    int n1=(g.vcnt+2)/3;
    int n2=(g.vcnt+1)/3;
    int n3=g.vcnt/3;
    
    int cnt1=0;
    rep(i,0,g.vcnt)if(dp[i])cnt1++;
    printf("%d %d\n",cnt1,n3);
    if(cnt1<=n3){
        rep(i,0,g.vcnt){
            if(dp[i]){
                printf("%d",n3--*3);
            }else{
                if(n1)printf("%d ",n1--*3-2);
                else if(n2)printf("%d ",n2--*3-1);
                else printf("%d ",n3--*3);
            }
        }
    }else if(g.vcnt-cnt1<=n3){
        rep(i,0,g.vcnt){
            if(!dp[i]){
                printf("%d",n3--*3);
            }else{
                if(n1)printf("%d ",n1--*3-2);
                else if(n2)printf("%d ",n2--*3-1);
                else printf("%d ",n3--*3);
            }
        }
    }else{
        rep(i,0,g.vcnt){
            if(dp[i]){
                if(n1)printf("%d ",n1--*3-2);
                else printf("%d ",n3--*3);
            }else{
                if(n2)printf("%d ",n2--*3-1);
                else printf("%d ",n3--*3);
            }
        }
    }
}