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>
#define ll long long
#define vi vector<int>
#define pii pair<int,int>
#define pb push_back

using namespace std;
const int N=2e6+5;
const int mod=1e9+7;

int n,val[N];
vi adj[N];

vi b,r; 
// bipartite

void dfs(int x,int p, int h=0){
    if(h%2) b.push_back(x);
    else r.push_back(x);
    for(auto xx: adj[x]) if(xx!=p){
        dfs(xx,x,h+1);
    }
}

vi c[3];
int fill(int i, int j, int k){
    int t;
    if(c[i].size()) t=c[i].back(),c[i].pop_back();
    else if(c[j].size()) t=c[j].back(),c[j].pop_back();
    else t=c[k].back(),c[k].pop_back();
    return t;
}

int main(){
    
    cin>>n;
    for(int i=1;i<n;i++) {
        int u,v;
        cin>>u>>v;
        adj[u].push_back(v);
        adj[v].push_back(u);
    }    

    for(int i=1;i<=n;i++) c[i%3].push_back(i);
    dfs(1,-1);

    if(b.size()<=n/3){
        for(auto x: b) val[x]=fill(0,1,2);
        for(auto x: r) val[x]=fill(1,2,0);
    }else if(r.size()<=n/3){
        for(auto x: r) val[x]=fill(0,1,2);
        for(auto x: b) val[x]=fill(1,2,0);
    }else{
        for(auto xx: b) val[xx]= fill(1,0,2);
        for(auto xx: r) val[xx]= fill(2,0,1);
    }
    for(int i=1;i<=n;i++) cout<<val[i]<<" ";
    return 0;
}