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<iostream>
#include<vector>
#include<map>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<string>
#include<algorithm>
#include<stack>
#include<queue>
#include<set>
#include<bitset>
#include<iomanip>
#define BUG puts("no bug")
#define sci(x) scanf("%d",&x)
#define scl(x) scanf("%lld",&x)
#define ll long long
#define ull unsigned long long
#define REP(a,b,c) for(int a=b;a<=c;a++)
#define PER(a,b,c) for(int a=b;a>=c;a--)
#define lb(a) (a&(-a))
#define pp (p<<1)
#define ppp (p<<1|1)
#define pi pair<int,int>
#define pii pair<pair<int,int>,int>
using namespace std;
const int MAXN=1e6+5;
int n,m,t;
int ans[MAXN];
vector<int>bian[MAXN];
vector<int>color[2];
void dfs(int x,int now,int fa){
    color[now].push_back(x);
    for(int i=0;i<bian[x].size();i++){
        int to=bian[x][i];
        if(to==fa)continue;
        dfs(to,now^1,x);
    }
}
void getans(int co,int &mod){
    while(color[co].size()&&mod<=n){
        ans[color[co].back()]=mod;
        color[co].pop_back();
        mod+=3;
    }
}
int main(){
    cin>>n;
    REP(i,1,n-1){
        int a,b;
        sci(a);sci(b);
        bian[a].push_back(b);
        bian[b].push_back(a);
    }
    dfs(1,0,0);
    int a=1,b=2,c=3;
    getans(color[0].size()<color[1].size(),a);
    getans(color[0].size()<color[1].size(),b);
    getans(1,c);
    getans(0,c);
    REP(i,1,n)cout<<ans[i]<<' ';
}