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;
#define mp make_pair
const ll mod=1e9+7;
const int maxn=2e5;
vector<int> g[maxn+5];
vector<int> tmp[2];
int ans[maxn+5];
int num[3];
int n;
void dfs(int k,int fa,int dep)
{
    tmp[dep&1].push_back(k);
    for(auto u:g[k])
        if(u!=fa)
            dfs(u,k,dep+1);
}
int main()
{

    ios::sync_with_stdio(false);cin.tie(0);
    cin>>n;
    num[0]=n/3;
    num[1]=n/3+(n%3>=1);
    num[2]=n/3+(n%3>=2);
    for(int i=1;i<n;++i)
    {
        int u,v;
        cin>>u>>v;
        g[u].push_back(v);
        g[v].push_back(u);
    }
    dfs(1,0,0);
    if(tmp[0].size()>tmp[1].size()) swap(tmp[0],tmp[1]);
    // cout<<tmp[0].size()<<" "<<tmp[1].size()<<endl;
    if(tmp[0].size()<=num[0])
    {
        for(auto x:tmp[0]) ans[x]=0;
        int i=0;
        for(;i<num[0]-int(tmp[0].size());++i) ans[tmp[1][i]]=0;
        for(;i<num[0]-int(tmp[0].size())+num[1];++i) ans[tmp[1][i]]=1;
        for(;i<tmp[1].size();++i) ans[tmp[1][i]]=2;
    }
    else
    {
        int i=0;
        for(;i<num[1];++i) ans[tmp[1][i]]=1;
        for(;i<tmp[1].size();++i) ans[tmp[1][i]]=0;
        i=0;
        for(;i<num[2];++i) ans[tmp[0][i]]=2;
        for(;i<tmp[0].size();++i) ans[tmp[0][i]]=0;
    }
    // for(int i=1;i<=n;++i) cout<<ans[i]<<" ";cout<<endl;
    num[0]=3;
    num[1]=1;
    num[2]=2;
    for(int i=1;i<=n;++i)
    {
        printf("%d ",num[ans[i]]);
        num[ans[i]]+=3;
    }
    return 0;
}