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 <algorithm>
#include <vector>
using namespace std;
int n;
vector< vector< int > > edg;
vector< int > c, p, sz, nx;

void f(int x, int y){
    c[x] = y;
    sz[y]++;
    for(int v : edg[x]) if(c[v] == -1) f(v, 1-y);
}

int get(int x){
    int res = 0;
    if(nx[x] <= n){
        res = nx[x];
        nx[x] += 3;
    }else{
        for(int i=0;;i++){
            if(nx[i] <= n){
                res = nx[i];
                nx[i] += 3;
                break;
            }
        }
    }
    return res;
}

int main(void){
    cin >> n;
    edg.assign(n, vector<int>());
    for(int i=0;i<n-1;i++){
        int a, b;
        cin >> a >> b;
        a--;
        b--;
        edg[a].push_back(b);
        edg[b].push_back(a);
    }
    c.assign(n, -1);
    p.assign(n, -1);
    sz.assign(2, 0);
    nx.assign(3, 3); nx[1] = 1; nx[2] = 2;
    f(0, 0);
    if(sz[0] <= n/3){
        for(int i=0;i<n;i++) if(c[i] == 0) p[i] = get(0);
    }else if(sz[1] <= n/3){
        for(int i=0;i<n;i++) if(c[i] == 1) p[i] = get(0);
    }else{
        for(int i=0;i<n;i++){
            if(c[i] == 0) p[i] = get(1);
            if(c[i] == 1) p[i] = get(2);
        }
    }
    for(int i=0;i<n;i++){
        if(p[i] == -1) p[i] = get(0);
        cout << p[i] << " ";
    }
    cout << endl;
}