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 rep(i,n) for(int i=0;i<n;++i)
using namespace std;
using ll=long long;
constexpr ll mod=1000000007;
constexpr int inf=1000000000;
constexpr ll linf=1000000000000000000;

vector<int> e[200000];
int num[200000];
int bl=0,wh=0;

void dfs(int v,int p,int t){
    if(t==1)++bl;
    else ++wh;
    num[v]=t;
    for(auto u:e[v]){
        if(u!=p){
            dfs(u,v,-t);
        }
    }
}

void print(int & a){
    cout<<a<<" ";
    a+=3;
}

int main(){
    int n;
    cin>>n;
    for(int i=0;i<n-1;i++){
        int s,t;
        cin>>s>>t;
        --s,--t;
        e[s].push_back(t);
        e[t].push_back(s);
    }
    dfs(0,-1,1);
    int a=1,b=2,c=3;
    if(bl<=n/3){
        rep(i,n){
            if(num[i]==1)print(c);
            else if(a<=n)print(a);
            else if(b<=n)print(b);
            else print(c);
        }
    }
    else if(wh<=n/3){
        rep(i,n){
            if(num[i]==-1)print(c);
            else if(a<=n)print(a);
            else if(b<=n)print(b);
            else print(c);
        }
    }
    else{
        rep(i,n){
            if(num[i]==1){
                if(a<=n)print(a);
                else print(c);
            }
            else{
                if(b<=n)print(b);
                else print(c);
            }
        }
    }
    cout<<endl;
}