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:
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
#include <bits/stdc++.h>
using namespace std;
template<class t> inline t read(t &x){
	char c=getchar();bool f=0;x=0;
	while(!isdigit(c)) f|=c=='-',c=getchar();
	while(isdigit(c)) x=(x<<1)+(x<<3)+(c^48),c=getchar();
	if(f) x=-x;return x;
}
template<class t,class ...A> inline void read(t &x,A &...a){
	read(x);read(a...);
}
template<class t> inline void write(t x){
	if(x<0) putchar('-'),write(-x);
	else{if(x>9) write(x/10);putchar('0'+x%10);}
}

const int N=2e5+5;
vector<int> g[N],d[2],a[3];
int n,ans[N];

void dfs(int x,int fa,int p){
	d[p].push_back(x);
	for(int y:g[x]) if(y^fa) dfs(y,x,p^1);
}

signed main(){
	read(n);
	for(int i=1;i<=n;i++) a[i%3].push_back(i);
	for(int i=1,x,y;i<n;i++){
		read(x,y);
		g[x].push_back(y);
		g[y].push_back(x);
	}
	dfs(1,0,0);
	if(d[0].size()>d[1].size()) swap(d[0],d[1]);
	if(a[1].size()>d[0].size()) for(int x:d[0]) ans[x]=a[0].back(),a[0].pop_back();
	else for(int x:d[0])
		if(a[1].size()) ans[x]=a[1].back(),a[1].pop_back();
		else ans[x]=a[0].back(),a[0].pop_back();
	for(int x:d[1])
		if(a[0].size()) ans[x]=a[0].back(),a[0].pop_back();
		else if(a[1].size()) ans[x]=a[1].back(),a[1].pop_back();
		else ans[x]=a[2].back(),a[2].pop_back();
	for(int i=1;i<=n;i++) write(ans[i]),putchar(' ');
}