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 pb push_back
using namespace std;
const int N=2e5+10;
int n,ans[N];bool vis[N];
vector<int>e[N],R,B;
void dfs(int x,int ff,int op){
    op?R.pb(x):B.pb(x);
    if(ff)e[x].erase(find(e[x].begin(),e[x].end(),ff));
    for(int i=0;i<e[x].size();i++)dfs(e[x][i],x,op^1);
}
int main(){
    scanf("%d",&n);
    for(int i=1,x,y;i<n;i++){
        scanf("%d%d",&x,&y);
        e[x].pb(y);e[y].pb(x);
    }
    dfs(1,0,0);
    if(R.size()>B.size())swap(R,B);
    if(R.size()<=n/3){
        for(int i=0,j=3;i<R.size();i++,j+=3)vis[ans[R[i]]=j]=1;
        for(int i=0,j=1;i<B.size();i++,j++){
            while(vis[j])j++;ans[B[i]]=j;
        }
    }
    else{
        int p=0,q=0;
        for(int i=1;i<=n;i+=3,p++)vis[ans[R[p]]=i]=1;
        for(int i=2;i<=n;i+=3,q++)vis[ans[B[q]]=i]=1;
        for(int i=1;p<R.size();p++,i++){
            while(vis[i])i++;vis[ans[R[p]]=i]=1;
        }
        for(int i=1;q<B.size();q++,i++){
            while(vis[i])i++;ans[B[q]]=i;
        }
    }
    for(int i=1;i<=n;i++)printf("%d ",ans[i]);puts("");
    return 0;
}