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;
#define rep(i,n) for(int i=0;i<n;i++)
typedef long long ll;
typedef pair<int,int> P;
typedef vector<int> vec;
typedef vector<vec> mat;
vec G[200001];
int V,color[200001];
bool dfs(int v,int c){
    color[v]=c;
    rep(i,G[v].size()){
        if(color[G[v][i]]==c) return false;
        if(color[G[v][i]]==0&&!dfs(G[v][i],-c)) return false;
    }
    return true;
}
int main(){
    int n,a,b;
    cin>>n;
    rep(i,n-1) cin>>a>>b,G[a].push_back(b),G[b].push_back(a);
    dfs(1,1);
    int x=0,y=0;
    rep(i,n){
        if(color[i+1]==1) x++;
        else y++;
    }
    int Z=-1;
    if(x<y) swap(x,y),Z=1;
    int X=(n+2)/3,Y=(n+1)/3;
    vec ans(200001);
    int xx=0,yy=0,zzz=0;
    if(y<Y){
        rep(i,n){
            if(color[i+1]==Z) ans[i+1]=3*zzz+3,zzz++;
            else{
                if(xx<X) ans[i+1]=3*xx+1,xx++;
                else if(yy<Y) ans[i+1]=3*yy+2,yy++;
                else ans[i+1]=3*zzz+3,zzz++;
            }
        }
        rep(i,n) cout<<ans[i+1]<<" ";
        return 0;
    }
    rep(i,n){
        if(color[i+1]==Z){
            if(yy<Y) ans[i+1]=3*yy+2,yy++;
            else ans[i+1]=3*zzz+3,zzz++;
        }
        else{
            if(xx<X) ans[i+1]=3*xx+1,xx++;
            else ans[i+1]=3*zzz+3,zzz++;
        }
    }
    rep(i,n) cout<<ans[i+1]<<" ";
}