Problem: 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.

Example

Input

5
1 2
1 3
3 4
3 5


Output

1 2 5 4 3

Solution (in CPP):

Solution:
#include <cstdio>
#include <vector>
#include <cstring>
#define pb push_back
#define catans(x) for(int i=1;i<=kn[x];++i)k[++pt]=(i-1)*3+x
using namespace std;vector<int> c[2];int id[200001],ans[200001],k[200001],n,pt,kn[4];
struct edge{int v,next;edge(){}edge(int _v,int _next){v=_v;next=_next;}}e[400001];int head[200001],sz;
void init(){memset(head,-1,sizeof(head));sz=0;}void insert(int u,int v){e[++sz]=edge(v,head[u]);head[u]=sz;}
void dfs(int u,int f,int d){c[d].pb(u);for(int i=head[u];~i;i=e[i].next)if(e[i].v!=f)dfs(e[i].v,u,d^1);}
int main()
{
	scanf("%d",&n);init();for(int i=1;i<n;++i){int u,v;scanf("%d%d",&u,&v);insert(u,v);insert(v,u);}
	dfs(1,0,0);if(c[0].size()>c[1].size())swap(c[0],c[1]);kn[3]=n/3;kn[1]=(n+2)/3;kn[2]=(n+1)/3;
	for(int i=1;i<=c[0].size();++i)id[i]=c[0][i-1];for(int i=1;i<=c[1].size();++i)id[i+c[0].size()]=c[1][i-1];
	if(n/3>=c[0].size()){catans(3);catans(1);catans(2);}else{catans(1);catans(3);catans(2);}
	for(int i=1;i<=n;++i)ans[id[i]]=k[i];for(int i=1;i<=n;++i)printf("%d ",ans[i]);putchar(10);return 0;
}