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;
typedef long long ll;
const ll N=2e5+5;
ll col[N],ans[N],cnt[2];
vector<ll> adj[N],pos[3];
void dfs(ll node,ll pr){
    col[node]=(col[pr]^1);
    cnt[col[node]]++;
    for(auto &u:adj[node])if(u!=pr)dfs(u,node);
}
int main(){
    ios_base:: sync_with_stdio(false);
    cin.tie(NULL); cout.tie(NULL);
    ll n;cin>>n;
    for(ll i=1;i<n;i++){
        ll a,b;cin>>a>>b;
        adj[a].push_back(b);
        adj[b].push_back(a);
    }
    for(ll i=1;i<=n;i++)pos[i%3].push_back(i);
    dfs(1,0);
    ll s=0;
    if(cnt[1]<cnt[0])s=1;

    if(cnt[s]<=pos[0].size()){
        for(ll i=1;i<=n;i++){
            if(col[i]==s){
                ans[i]=pos[0].back();
                pos[0].pop_back();
            }
        }
    }
    else{
        for(ll i=1;i<=n;i++){
            if(col[i]==s){
                if(!pos[s+1].empty()){
                    ans[i]=pos[s+1].back();
                    pos[s+1].pop_back();
                }
                else{
                    ans[i]=pos[0].back();
                    pos[0].pop_back();
                }
            }
        }
    }

    for(ll i=1;i<=n;i++){
        if(!ans[i]){
            for(ll j=0;j<3;j++){
                if(!pos[j].empty()){
                    ans[i]=pos[j].back();
                    pos[j].pop_back();
                    break;
                }
            }
        }
    }

    for(ll i=1;i<=n;i++)cout<<ans[i]<<" ";
}