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 maxn = 200010;
int n, ans[maxn];
bool mark[maxn];
vector<int> G[maxn], V[2];

int main() {
    scanf("%d", &n);
    for (int i = 1, u, v; i < n; i++) {
        scanf("%d %d", &u, &v);
        G[u].push_back(v), G[v].push_back(u);
    }
    function<void(int, int, int)> dfs = [&](int v, int fa, int d) {
        V[d].push_back(v);
        for (int u : G[v]) if (u ^ fa) dfs(u, v, d ^ 1);
    };
    dfs(1, 0, 0);
    if (V[0].size() > V[1].size()) swap(V[0], V[1]);
    if (V[0].size() <= n / 3) {
        for (int i = 0; i < V[0].size(); i++) {
            ans[V[0][i]] = (i + 1) * 3, mark[(i + 1) * 3] = 1;
        }
    } else {
        int t = min((n + 2) / 3, (int)V[0].size());
        for (int i = 0; i < t; i++) {
            ans[V[0][i]] = i * 3 + 1, mark[i * 3 + 1] = 1;
        }
        for (int i = t; i < V[0].size(); i++) {
            ans[V[0][i]] = (i - t + 1) * 3, mark[(i - t + 1) * 3] = 1;
        }
    }
    int cur = 1;
    for (int i = 1; i <= n; i++) if (!ans[i]) {
        while (mark[cur]) cur++;
        ans[i] = cur, mark[cur] = 1;
    }
    for (int i = 1; i <= n; i++) {
        printf("%d ", ans[i]);
    }
    return 0;
}