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>
#pragma comment(linker, "/STACK:16000000")
using namespace std;

const int Maxn = 200005;

int n;
vector <int> neigh[Maxn];
vector <int> inds[2];
vector <int> vals[3];
int res[Maxn];

void Traverse(int v, int p, int lvl)
{
    vals[v % 3].push_back(v);
    inds[lvl % 2].push_back(v);
    for (int i = 0; i < neigh[v].size(); i++) {
        int u = neigh[v][i];
        if (u == p) continue;
        Traverse(u, v, lvl + 1);
    }
}

void Assign(vector <int> &inds, vector <int> &vals)
{
    while (!inds.empty() && !vals.empty()) {
        res[inds.back()] = vals.back();
        inds.pop_back();
        vals.pop_back();
    }
}

int main()
{
    scanf("%d", &n);
    for (int i = 0; i < n - 1; i++) {
        int a, b; scanf("%d %d", &a, &b);
        neigh[a].push_back(b);
        neigh[b].push_back(a);
    }
    Traverse(1, 0, 0);
    if (inds[0].size() > inds[1].size())
        swap(inds[0], inds[1]);
    if (vals[0].size() >= inds[0].size()) {
        Assign(inds[0], vals[0]);
        Assign(inds[1], vals[0]);
        Assign(inds[1], vals[1]);
        Assign(inds[1], vals[2]);
    } else {
        Assign(inds[0], vals[1]);
        Assign(inds[0], vals[0]);
        Assign(inds[1], vals[0]);
        Assign(inds[1], vals[2]);
    }
    for (int i = 1; i <= n; i++)
        printf("%d%c", res[i], i + 1 <= n? ' ': '\n');
    return 0;
}