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 N 200002
using namespace std;
vector<int> v[N],vv[2],nr[3];
int c[N],l[N];
void dfs(int i)
{
    vv[l[i]%2].push_back(i);
    for(auto it:v[i])
        if(!l[it])
            l[it]=l[i]+1,dfs(it);
}
void ad(int i,int j)
{
    c[vv[i].back()]=nr[j].back();
    vv[i].pop_back();
    nr[j].pop_back();
}
int main()
{
    int n,i,j,k;
    cin>>n;
    for(k=1;k<n;k++)
    {
        cin>>i>>j;
        v[i].push_back(j);
        v[j].push_back(i);
        nr[k%3].push_back(k);
    }
    nr[n%3].push_back(n);
    l[1]=1;
    dfs(1);
    if(vv[0].size()<=n/3)
    {
        while(vv[0].size())
            ad(0,0);
        while(vv[1].size())
            ad(1,nr[0].size()?0:(nr[1].size()?1:2));
    }
    else if(vv[1].size()<=n/3)
    {
        while(vv[1].size())
            ad(1,0);
        while(vv[0].size())
            ad(0,nr[0].size()?0:(nr[1].size()?1:2));
    }
    else
    {
        while(vv[0].size())
            ad(0,nr[1].size()?1:0);
        while(vv[1].size())
            ad(1,nr[0].size()?0:2);
    }
    for(i=1;i<=n;i++)
        cout<<c[i]<<" ";
    return 0;
}