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 rep(i,l,r) for (int i=l,i##end=r;i<=i##end;++i)
#define per(i,r,l) for (int i=r,i##end=l;i>=i##end;--i)
#define pb push_back
using namespace std;
const int N=2e5+5;
vector<int> G[N],a[3];
int ans[N];
void dfs(int u,int f,int d){
    a[d].pb(u);
    rep(i,0,G[u].size()-1){ int v=G[u][i]; if (v!=f) dfs(v,u,d^1); }
}
int main(){
    int n; scanf("%d",&n);
    rep(i,1,n-1){ int u,v; scanf("%d%d",&u,&v); G[u].pb(v),G[v].pb(u); }
    dfs(1,0,1);
    if (a[0].size()>a[1].size()) swap(a[0],a[1]);
    if (a[0].size()<=n/3){
        rep(i,0,a[0].size()-1) ans[a[0][i]]=(i+1)*3;
        int c=0;
        rep(i,1,n){
            if (i%3==0) continue;
            ans[a[1][c++]]=i;
        }
        c=a[0].size();
        rep(i,1,n) if (!ans[i]) ans[i]=(++c)*3;
    }
    else{
        int c=0;
        rep(i,0,a[0].size()-1)
            if (i*3+2<=n) ans[a[0][i]]=i*3+2;
            else ans[a[0][i]]=(++c)*3;
        rep(i,0,a[1].size()-1)
            if (i*3+1<=n) ans[a[1][i]]=i*3+1;
            else ans[a[1][i]]=(++c)*3;
    }
    rep(i,1,n) printf("%d ",ans[i]);
    return 0;
}