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;
int n;
vector<int>edge[200005];
vector<int>vt[2], num[3];
int col[200005], ans[200005];

int main(){
	cin >> n;
	for(int i=0;i<n-1;i++){
		int a,b; cin >> a >> b;
		edge[a].emplace_back(b);
		edge[b].emplace_back(a);
	}
	function<void(int,int,int)> dfs = [&](int v,int u,int c){
		col[v] = c; vt[c].emplace_back(v);
		for(auto at:edge[v]) if(at != u) dfs(at,v,c^1);
	};
	dfs(1,-1,0);
	
	for(int i=1;i<=n;i++) num[i%3].emplace_back(i);
	for(int mask=0;mask<4;mask++){
		int x = 0, y = 0;
		for(int i=0;i<2;i++) if(((mask>>i)&1)) x += num[1+i].size();
		y = n - x - num[0].size();
		
		if(x <= vt[0].size() && y <= vt[1].size()){
			int nxt,sz = 0;
			nxt = 0;
			for(int i=0;i<2;i++) if(((mask>>i)&1)) {
				for(auto at:num[1+i]) ans[vt[0][nxt++]] = at;
			}
			while(nxt < vt[0].size()) ans[vt[0][nxt++]] = num[0][sz++];
			
			nxt = 0;
			for(int i=0;i<2;i++) if(!((mask>>i)&1)) {
				for(auto at:num[1+i]) ans[vt[1][nxt++]] = at;
			}
			while(nxt < vt[1].size()) ans[vt[1][nxt++]] = num[0][sz++];
			assert(sz == num[0].size());
			for(int i=1;i<=n;i++) printf("%d ",ans[i]);
			puts(""); return 0;
		}
	}
	puts("-1");
}