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;
const int N = 200005;
int ans[N], col[N];
vector<int> adjlist[N];
vector<int> sets[2];
void dfs(int u, int p){
    sets[col[u]].push_back(u);
    for (auto v : adjlist[u]){
        if (v == p) continue;
        col[v] = 1-col[u];
        dfs(v,u);
    }
}
int main(){
    int n;
    scanf("%d",&n);
    for (int i = 1; i < n; i++){
        int a,b;
        scanf("%d%d",&a,&b);
        adjlist[a].push_back(b);
        adjlist[b].push_back(a);
    }
    col[1] = 1;
    dfs(1,-1);
    if (sets[0].size() > sets[1].size()) swap(sets[0],sets[1]);
    int X = n/3;
    int R = sets[0].size(), B = sets[1].size();
    int lim1 = (n+2)/3, lim2 = (n+1)/3;
    if (R <= X){
        for (auto x : sets[0]) ans[x] = 3;
        for (auto x : sets[1]){
            if (lim1) {
                lim1--;
                ans[x] = 1;
            }
            else if (lim2){
                lim2--;
                ans[x] = 2;
            }
            else ans[x] = 3;
        }
    }
    else{
        for (auto x : sets[0]){
            if (lim1) {
                lim1--;
                ans[x] = 1;
            }
            else ans[x] = 3;
        }
        for (auto x : sets[1]){
            if (lim2) {
                lim2--;
                ans[x] = 2;
            }
            else ans[x] = 3;
        }
    }
    int cur[4]; iota(cur,cur+4,0);
    for (int i = 1; i <= n; i++){
        printf("%d ",cur[ans[i]]);
        cur[ans[i]]+=3;
    }
}