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 int ll
typedef long double ld;

typedef pair<int,int> pii;
#define x first
#define y second

const int N=2e5+100;

vector<int> g[N];
vector<int> part[2];
int p[N];

void dfs(int v,int p,int d)
{
    part[d%2].push_back(v);
    for(int to:g[v])
        if(to!=p)
            dfs(to,v,d+1);
}

signed main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);

    int n;
    cin>>n;
    for(int i=1;i<=n-1;i++)
    {
        int a,b;
        cin>>a>>b;
        g[a].push_back(b);
        g[b].push_back(a);
    }
    dfs(1,-1,0);
    if(part[0].size()>part[1].size())
        swap(part[0],part[1]);
    int cnt[3]={};
    for(int i=1;i<=n;i++)
        cnt[i%3]++;
    vector<int> ord;
    if(part[0].size()<=cnt[0])
    {
        for(int i=1;i<=n;i++)
            if(i%3==0)
                ord.push_back(i);
        for(int i=1;i<=n;i++)
            if(i%3!=0)
                ord.push_back(i);
    }
    else
    {
        for(int i=1;i<=n;i++)
            if(i%3==1)
                ord.push_back(i);
        for(int i=1;i<=n;i++)
            if(i%3==0)
                ord.push_back(i);
        for(int i=1;i<=n;i++)
            if(i%3==2)
                ord.push_back(i);
    }
    vector<int> vers=part[0];
    copy(part[1].begin(),part[1].end(),back_inserter(vers));
    for(int i=0;i<n;i++)
        p[vers[i]]=ord[i];
    for(int i=1;i<=n;i++)
        cout<<p[i]<<" ";
}