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 rep(i, n) for (int i = 0; i < (int)(n); i++)
typedef int64_t Int;
using namespace std;

int N; 
vector<vector<int>> G(200010);
vector<int> parity(200010, -1);

vector<int> ans(200010, -1);
vector<bool> used(2000010, false);

void fill(){
    int id = 0;
    for(int i=1; i<=N; i++){
        if(used[i]) continue;

        while(ans[id] != -1) id++;
        ans[id] = i;
    }
}

void bfs(){
    queue<int> Q; Q.push(0); parity[0] = 0;
    bool E = true;
    while(!Q.empty()){
        int v = Q.front(); Q.pop();

        rep(i, G[v].size()){
            int u = G[v][i];
            if(parity[u] == -1){
                parity[u] = 1-parity[v];
                Q.push(u);
            }
        }
    }
}
 
int main(){
    cin >> N;
    rep(i, N-1){
        int a, b; cin >> a >> b; a--; b--;
        G[a].push_back(b); G[b].push_back(a);
    }

    bfs();

    vector<int> even, odd;
    rep(i, N){
        if(parity[i] == 0) even.push_back(i);
        else odd.push_back(i);
    }
    int E = even.size(), O = odd.size();

    
    if(E > N/3 && O > N/3){
        rep(i, E) if(3*i+1 <= N) {ans[even[i]] = 3*i+1; used[3*i+1] = true;}
        rep(i, O) if(3*i+2 <= N) {ans[odd[i]] = 3*i+2; used[3*i+2] = true;}
        fill();
    } else if(E <= N/3){
        rep(i, E) {ans[even[i]] = 3*(i+1); used[3*(i+1)] = true;}
        fill();
    } else if(O <= N/3){
        rep(i, O) {ans[odd[i]] = 3*(i+1); used[3*(i+1)] = true;}
        fill();
    }

    rep(i, N) cout << ans[i] << " ";
    cout << endl;
}