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 <cstdio>
#include <algorithm>
#include <queue>
#include <cmath>
#include <cstring>
#include <vector>
#include <map>
using namespace std;
typedef long long ll;
const int N = 2e5 + 10;
const int mod = 1e9 + 7;

vector<int> graf[N];
vector<int> h[2];
void dfs(int x, int y = 0, int d = 0)
{
    h[d].push_back(x);
    for (int v : graf[x])
        if (v != y)
            dfs(v, x, d ^ 1);
}
int ans[N];
int main()
{
    int n;
    scanf("%d", &n);
    for(int i=1;i<n;i++)
    {
        int a, b;
        scanf("%d%d", &a, &b);
        graf[a].push_back(b);
        graf[b].push_back(a);
    }
    dfs(1, 0);
    vector<int> p[3];
    for (int i = 1; i <= n; ++i)
        p[i % 3].push_back(i);

    for (int c = 1; c <= 2; ++c)
    {
        bool did = true;
        for (int d =0; d < 2;d++)
            if (h[d].size() >= p[c].size())
            {
                for (int x : p[c])
                {
                    ans[h[d].back()] = x;
                    h[d].pop_back();
                }
                did = true;
                break;
            }
        //assert(did);
    }
    for (int i = 1; i <= n; ++i)
        if (ans[i] == 0)
        {
            ans[i] = p[0].back();
            p[0].pop_back();
        }
    for (int i = 1; i <= n; ++i)
        printf("%d ", ans[i]);
    printf("\n");
}