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:
// editorial
// 思いつける気がしない
#include<iostream>
#include<vector>
#include<queue>
using namespace std;

vector<int> v[2];
vector<int> g[200000];

void dfs(int x, int p, int k){
    v[k].push_back(x);
    for(int next : g[x]){
        if(next == p)   continue;
        dfs(next, x, 1-k);
    }
}

int main(){
    int n;
    cin >> n;
    for(int i = 0; i < n-1; i++){
        int a, b;
        cin >> a >> b;
        a--, b--;
        g[a].push_back(b);
        g[b].push_back(a);
    }
    dfs(0, -1, 0);
    if(v[0].size() > v[1].size())   swap(v[0], v[1]);
    vector<int> ans(n, 0);
    queue<int> q[3];
    for(int i = 1; i <= n; i++) q[i%3].push(i);
    if(v[0].size() >= q[1].size()){
        for(int x : v[0]){
            if(q[1].empty())    ans[x] = q[0].front(), q[0].pop();
            else                ans[x] = q[1].front(), q[1].pop();
        }
        for(int x : v[1]){
            if(q[2].empty())    ans[x] = q[0].front(), q[0].pop();
            else                ans[x] = q[2].front(), q[2].pop();
        }
    }else{
        for(int x : v[0])   ans[x] = q[0].front(), q[0].pop();
        for(int x : v[1]){
            for(int i = 0; i < 3; i++){
                if(q[i].empty())    continue;
                ans[x] = q[i].front();  q[i].pop();
                break;
            }
        }
    }
    for(int i = 0; i < n; i++)  cout << ans[i] << " \n"[i==n-1];
    return 0;
}