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<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef vector<int> vi;
typedef pair<int,int> pii;
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define sz(a) int(a.size())
const int N=2e5+10;

int gi() {
	int x=0,o=1;char ch=getchar();
	while((ch<'0'||ch>'9')&&ch!='-') ch=getchar();
	if(ch=='-') o=-1,ch=getchar();
	while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
	return x*o;
}

int n,fa[N],cnt[2],ans[N];
vi E[N],vec[3];
bool col[N];

void dfs(int u,int ff,bool col) {
	fa[u]=ff;::col[u]=col;++cnt[col];
	for(auto v:E[u]) if(v^ff) dfs(v,u,col^1);
}

void out(int y) {
	cout<<vec[y].back()<<' ';vec[y].pop_back();
}

void out2(int x,int y) {
	ans[x]=vec[y].back();vec[y].pop_back();
}

int main() {
	cin>>n;
	for(int i=1,u,v;i<n;i++) u=gi(),v=gi(),E[u].pb(v),E[v].pb(u);
	dfs(1,0,0);
	for(int i=1;i<=n;i++) vec[i%3].pb(i);
	if(cnt[0]>cnt[1]) {
		swap(cnt[0],cnt[1]);
		for(int i=1;i<=n;i++) col[i]^=1;
	}
	if(cnt[0]<=sz(vec[0])) {
		for(int i=1;i<=n;i++)
			if(col[i]==0) out2(i,0);
		for(int i=1;i<=n;i++)
			if(col[i]==1) {
				for(int j=0;j<3;j++) if(!vec[j].empty()) { out2(i,j);break; }
			}
		for(int i=1;i<=n;i++) cout<<ans[i]<<' ';
	}
	else {
		for(int i=1;i<=n;i++) {
			if(col[i]==0) {
				if(!vec[1].empty()) out(1);
				else out(0);
			}
			else {
				if(!vec[2].empty()) out(2);
				else out(0);
			}
		}
	}
	return 0;
}