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 = 2e5 + 5;
vector<int> E[maxn], G[3], V[2];
int T[maxn], n, Col[maxn], chk[maxn], Res[maxn];

void dfs(int a, int pp) {
    Col[a] = !Col[pp];
    V[Col[a]].push_back(a);
    for (auto &x : E[a]) {
        if (x == pp) continue;
        dfs(x, a);
    }
}

int main() {
    int i, a, b;
    scanf("%d", &n);
    for (i = 1; i <= n; ++i) {
        G[i % 3].push_back(i);
    }
    for (i = 1; i < n; ++i) {
        scanf("%d %d", &a, &b);
        E[a].push_back(b);
        E[b].push_back(a);
    }
    dfs(1, 0);
    int pv = 0;
    if (V[pv].size() > V[!pv].size()) pv = !pv;
    if (V[pv].size() <= G[0].size()) {
        int pp = 0;
        for (auto &x : V[pv]) {
            Res[x] = G[0][pp];
            chk[G[0][pp]] = 1;
            pp++;
        }
    }else {
        int pp = 0, pp2 = 0;
        for (auto &x : V[pv]) {
            if (pp < G[1].size()) {
                Res[x] = G[1][pp];
                chk[G[1][pp]] = 1;
                pp++;
            }else {
                Res[x] = G[0][pp2];
                chk[G[0][pp2]] = 1;
                pp2++;
            }
        }
    }
    vector<int> TP;
    for (i = 1; i <= n; ++i) {
        if (!chk[i]) {
            TP.push_back(i);
        }
    }
    for (i = 0; i < V[!pv].size(); ++i) {
        Res[V[!pv][i]] = TP[i];
    }
    for (i = 1; i <= n; ++i) {
        printf("%d ", Res[i]);
    }
    return 0;
}