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 fo(i,a,b) for(int i=(a);i<=(b);++i)
#define rv(i,a,b) for(int i=(a);i>=(b);--i)
#define SZ(x) ((int)x.size())
using namespace std;
const int M = 420000;
typedef long long ll;
int n,m,k;
vector<int> g[M];
vector<int> p[2];
int ans[M];
void dfs(int u,int c,int f){
    p[c].push_back(u);
    for(int to:g[u]){
        if(to==f) continue;
        dfs(to,c^1,u);
    }
}
int t[3];
int num(int x){
    if(x==0) return t[0]--*3;
    return --t[x]*3+x;
}
int num(){
    if(t[1]) return num(1);
    if(t[0]) return num(0);
    if(t[2]) return num(2);
    return -1;
}
int main(){
    cin>>n;
    int u,v;
    fo(i,1,n-1){
        cin>>u>>v;
        g[u].push_back(v);
        g[v].push_back(u);
    }
    dfs(1,0,1);
    int b=0,w=1;
    if(SZ(p[b])>SZ(p[w]))
        swap(b,w);
    t[0]=n/3;t[1]=(n+2)/3;t[2]=(n+1)/3;
    if(SZ(p[b])<=t[0]){
        for(int val:p[b])
            ans[val]=num(0);
        for(int val:p[w])
            ans[val]=num();
    }
    else{
        for(int val:p[b])
            ans[val]=num();
        for(int val:p[w])
            ans[val]=num();
    }
    fo(i,1,n) cout<<ans[i]<<' ';
    cout<<endl;
    return 0;
}