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>
using namespace std;
const int maxn=200005,maxe=maxn<<1;
int n,a,b,c,ans[maxn];
int e,lnk[maxn],nxt[maxe],son[maxe];
inline char nc(){
	static char buf[100000],*p1=buf,*p2=buf;
	return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;
}
inline int read(){
	int ret=0;bool f=0;char ch=nc();
	while(ch>'9'||ch<'0') f^=ch=='-',ch=nc();
	while(ch<='9'&&ch>='0') ret=ret*10+ch-'0',ch=nc();
	return f?-ret:ret;
}
inline void add_e(int x,int y){son[++e]=y;nxt[e]=lnk[x];lnk[x]=e;}
void DFS(int x,int fa,int y){
	if(y==0) if(a<=n) ans[x]=a,a+=3;else
			 if(b<=n) ans[x]=b,b+=3;else ans[x]=c,c+=3;
	if(y==1) if(a<=n) ans[x]=a,a+=3;else ans[x]=c,c+=3;
	if(y==2) if(b<=n) ans[x]=b,b+=3;else ans[x]=c,c+=3;
	if(y==3) ans[x]=c,c+=3;
	for (int i=lnk[x];i;i=nxt[i])
	if(son[i]^fa) DFS(son[i],x,y^3);
}
int main(){
	n=read();
	for (int i=1;i<n;i++){
		int x=read(),y=read();
		add_e(x,y),add_e(y,x);
	}
	a=1,b=2,c=3;DFS(1,-1,1);
	if(a<=n) a=1,b=2,c=3,DFS(1,-1,3);else
	if(b<=n) a=1,b=2,c=3,DFS(1,-1,0);
	for (int i=1;i<=n;i++) printf("%d ",ans[i]);
	return 0;
}