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;
using ll  = long long;
using vl  = vector<ll>;
using vvl = vector<vl>;
#define rng(i,a,b) for(int i = (a); i < (b); ++i)
#define rep(i,n) for(int i = 0; i < (n); ++i)

int N;
vvl edges;
vl color;
vl ans;

void input() {
    cin >> N;
    edges.resize(N);
    rep(i,N-1) {
        int a, b; cin >> a >> b;
        a--; b--;
        edges[a].push_back(b);
        edges[b].push_back(a);
    }
}

void dfs(int v, int pre, int c) {
    color[v] = c;
    c = 1 - c;
    for (auto to : edges[v]) {
        if (to == pre) continue;
        dfs(to, v, c);
    }
}
void coloring() {
    color.assign(N,0);
    dfs(0, -1, 0);
    // rep(i,N) cerr << color[i] << endl;
}

void solve() {
    coloring();

    vector<queue<int>> wannaColor(2);
    rep(i,N) wannaColor[color[i]].push(i);

    vector<queue<int>> restNums(3);
    rng(i,1,N+1) restNums[i%3].push(i);

    if (wannaColor[0].size() > wannaColor[1].size()) {
        swap(wannaColor[0], wannaColor[1]);
    }
    ans.assign(N,0);
    auto f = [&](int x, int y) {
        while (!wannaColor[x].empty() and !restNums[y].empty()) {
            ans[wannaColor[x].front()] = restNums[y].front();
            wannaColor[x].pop(); restNums[y].pop();
        }
    };
    if (wannaColor[0].size() <= N / 3) {
        f(0,0);
        f(1,0); f(1,1); f(1,2);
    } else {
        f(0,1); f(1,2);
        f(0,0); f(1,0);
    }
}

void output() {
    rep(i,N) cout << ans[i] << endl;
}

int main() {
    input();
    solve();
    output();
    return 0;
}