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 N=200005;
vector<int> v[N];
vector<int> col(N,0);
vector<int> x[2];

void dfs(int a,int p=0)
{
    col[a]=1-col[p];
    x[col[a]].push_back(a);
    for(int to:v[a])
    {
        if(to==p) continue;
        dfs(to,a);
    }
}

int main()
{
    int n;
    scanf("%d",&n);
    for(int i=0;i<n-1;i++)
    {
        int a,b;
        scanf("%d%d",&a,&b);
        v[a].push_back(b);
        v[b].push_back(a);
    }
    dfs(1);
    int one=(n+2)/3;
    int two=(n+1)/3;
    vector<int> p(n+1,0);
    if(x[0].size()<x[1].size()) swap(x[0],x[1]);
    for(int i=0;i<one;i++) p[x[0][i]]=1;
    if((int)x[1].size()>=two) for(int i=0;i<two;i++) p[x[1][i]]=2;
    else for(int i=one;i<one+two;i++) p[x[0][i]]=2;
    vector<int> t={0,-2,-1};
    for(int i=1;i<=n;i++)
    {
        printf("%d%c",t[p[i]]+=3," \n"[i==n]);
    }
    /*cout << "one=" << one << endl;
    cout << "two=" << two << endl;
    cout << "white: " << endl;
    for(int y:x[0]) cout << y << " ";
    cout << endl;
    cout << "black: " << endl;
    for(int y:x[1]) cout << y << " ";
    cout << endl;*/
    return 0;
}