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 <iostream>
#include <cstring>
#include <algorithm>
#include <vector>
#define PB push_back
#define MP make_pair
#define PII pair<int,int>
#define FIR first
#define SEC second
#define ll long long
using namespace std;
template <class T>
inline void rd(T &x) {
	x=0; char c=getchar(); int f=1;
	while(!isdigit(c)) { if(c=='-') f=-1; c=getchar(); }
	while(isdigit(c)) x=x*10-'0'+c,c=getchar(); x*=f;
}
const int N=2e5+10;
vector<int> d[2];
vector<int> G[N];
int dep[N],n;
void dfs(int u,int last) {
	dep[u]=dep[last]+1;
	d[dep[u]%2].PB(u);
	for(int v : G[u]) if(v!=last)
		dfs(v,u);
}
int ans[N],vis[N];
void gao() {
	int j=1;
	for(int i=1;i<=n;++i) if(!ans[i]) {
		while(vis[j]) ++j;
		ans[i]=j;
		++j;
	}
}
int main() {
	rd(n);
	for(int i=1,x,y;i<n;++i) rd(x),rd(y),G[x].PB(y),G[y].PB(x);
	dfs(1,0);
	if(d[0].size()>d[1].size()) swap(d[0],d[1]);
	if(d[0].size()<=n/3) {
		int v=3;
		for(int u : d[0])
			vis[ans[u]=v]=1,v+=3;
	} else {
		int v=1;
		for(int u : d[0]) {
			vis[ans[u]=v]=1,v+=3;
			if(v>n) break;
		}
		v=2;
		for(int u : d[1]) {
			vis[ans[u]=v]=1,v+=3;
			if(v>n) break;
		}
	}
	gao();
	for(int i=1;i<=n;++i) printf("%d ",ans[i]);
	return 0;
}