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<iostream>
#include<vector>
#include<algorithm>
#define ep emplace
#define eb emplace_back
#define fi first
#define se second
#define all(x) (x).begin(),(x).end()
using namespace std;
typedef long long ll;
typedef pair<int,int>pi;
typedef pair<ll,ll>pl;
const int inf=1e9+7;
const ll INF=1e18+7;
vector<int>adj[200010];
vector<int>tp[2];
void dfs(int x,int p,int t)
{
    tp[t].eb(x);
    for(int&v:adj[x])
        if(v!=p)
            dfs(v,x,t^1);
    return;
}
vector<int>mod[3];
int col[200010];
int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    int n;
    cin>>n;
    for(int i=1;i<n;i++)
    {
        int u,v;
        cin>>u>>v;
        adj[u].eb(v);
        adj[v].eb(u);
    }
    dfs(1,0,0);
    if(tp[0].size()>tp[1].size())
        swap(tp[0],tp[1]);
    for(int i=0;i++<n;)
        mod[i%3].eb(i);
    vector<int>ver;
    vector<int>cv;
    for(int&t:tp[0])
        ver.eb(t);
    for(int&t:tp[1])
        ver.eb(t);
    if(tp[0].size()>=mod[2].size())
    {
        for(int&t:mod[2])
            cv.eb(t);
        for(int&t:mod[0])
            cv.eb(t);
        for(int&t:mod[1])
            cv.eb(t);
    }
    else
    {
        for(int&t:mod[0])
            cv.eb(t);
        for(int&t:mod[1])
            cv.eb(t);
        for(int&t:mod[2])
            cv.eb(t);
    }
    for(int i=0;i<n;i++)
        col[ver[i]]=cv[i];
    for(int i=0;i++<n;)
        cout<<col[i]<<' ';
    cout<<endl;
    return 0;
}