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;
const int MAXN=200005;
vector<int> e[MAXN],col[2];
int dep[MAXN],usd[MAXN],res[MAXN];
void dfs(int u,int fa)
{
    dep[u]=dep[fa]+1;
    col[dep[u]&1].push_back(u);
    for(auto &v:e[u])
        if(v!=fa)dfs(v,u);
}
int main()
{
    int n;
    scanf("%d",&n);
    for(int i=1;i<n;i++)
    {
        int a,b;
        scanf("%d%d",&a,&b);
        e[a].push_back(b);
        e[b].push_back(a);
    }
    dfs(1,0);
    if(col[0].size()>col[1].size())
        swap(col[0],col[1]);
    if(col[0].size()<=n/3)
    {
        for(size_t i=0;i<col[0].size();i++)
            usd[res[col[0][i]]=3*i+3]=1;
        for(int i=1,j=0;i<=n;i++)
            if(!usd[i])res[col[1][j++]]=i;
    }
    else
    {
        int k=0;
        for(size_t i=0;i<(n+2)/3;i++)
            res[col[0][i]]=3*i+1;
        for(size_t i=(n+2)/3;i<col[0].size();i++)
            res[col[0][i]]=3*(++k);
        for(size_t i=0;i<(n+1)/3;i++)
            res[col[1][i]]=3*i+2;
        for(size_t i=(n+1)/3;i<col[1].size();i++)
            res[col[1][i]]=3*(++k);
    }
    for(int i=1;i<=n;i++)
        printf("%d%c",res[i]," \n"[i==n]);
    return 0;
}