Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

5
1 2
1 3
3 4
3 5

Sample Output 1

1 2 5 4 3

The distance between two vertices is 3 for the two pairs (2, 4) and (2, 5).

p_2 + p_4 = 6

p_2\times p_5 = 6

Thus, this permutation satisfies the condition.

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;

typedef long long LL;
typedef pair<int,int> pii;

#define pb push_back
#define mp make_pair

const int N = 2e5 + 10;

int n, ans[N], cur[3], fa[N];
vector<int> V[N];
bool g[N];
int dp[N];

void dfs(int x, int fa){
    dp[x] = 1;
    for(int i=0; i<V[x].size(); i++){
        int j = V[x][i];
        if(j == fa) continue;
        dfs(j, x);
        dp[x] += dp[j];
    }
}

pii dfs1(int x, int fa){
    pii ret = mp(n+1, 0);
    int m = 0;
    for(int i=0; i<V[x].size(); i++){
        int j = V[x][i];
        if(j == fa) continue;
        ret = min(ret, dfs1(j, x));
        m = max(m, dp[j]);
    }
    m = max(m, n - dp[x]);
    return min(ret, mp(m, x));
}

bool cmp(const int &a, const int &b){
    return dp[a] < dp[b];
}

void solve(int s){
    if(n > 2){
        ans[s] = 3;
        cur[0] += 3;
        g[s] = 0;
    } else {
        ans[s] = 1;
        cur[1] += 3;
        g[s] = 1;
    }
    queue<int> Q;
    Q.push(s);
    fa[s] = -1;

    int x;
    while(!Q.empty()){
        x = Q.front(); Q.pop();
        sort(V[x].begin(), V[x].end(), cmp);
        for(int i=0; i<V[x].size(); i++){
            int j = V[x][i];
            if(j == fa[x])  continue;
            fa[j] = x;
            if(g[x] == 1){
                ans[j] = cur[2];
                cur[2] += 3;
            } else {
                ans[j] = cur[1];
                cur[1] += 3;
            }
            g[j] = g[x] ^ 1;
            if(ans[j] > n){
                if(cur[0] <= n){
                    ans[j] = cur[0];
                    cur[0] += 3;
                } else {
                    if(g[j]){
                        ans[j] = cur[2];
                        cur[2] += 3;
                    } else {
                        ans[j] = cur[1];
                        cur[1] += 3;
                    }
                }
            }
            Q.push(j);
        }
    }
}

int main(){
    int x, y;
    scanf("%d", &n);
    for(int i=1; i<n; i++){
        scanf("%d %d", &x, &y);
        V[x].pb(y);
        V[y].pb(x);
    }
    cur[0] = 3;
    cur[1] = 1;
    cur[2] = 2;

    dfs(1, -1);
    int s = dfs1(1, -1).second;
    dfs(s, -1);

    solve(s);

    for(int i=1; i<=n; i++) printf("%d%c", ans[i], i==n?'\n':' ');

    return 0;
}