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;
#define rep(i,n) for(int i=0;i<int(n);++i)

vector<vector<int>> g;
vector<int> st[3];

void coloring(int cur, int pre, bool col){
    // col = 0 or 1
    st[col].push_back(cur);
    for(int to : g[cur]) {
        if (to == pre) continue;
        coloring(to, cur, !col);
    }
}

int main(void){
    int n;
    cin >> n;
    g.resize(n);
    rep(i,n-1){
        int a, b;
        cin >> a >> b;
        a--, b--;
        g[a].push_back(b);
        g[b].push_back(a);
    }
    coloring(0, -1, true);

    if (st[0].size() < st[1].size()) {
        swap(st[0], st[1]);
    }

    if ((int)st[1].size() < (n+1)/3) {
        swap(st[1], st[2]);
        rep(i,(n+1)/3){
            st[1].push_back(st[0].back());
            st[0].pop_back();
        }
    }
    while((int)st[0].size() > (n+2)/3) {
        st[2].push_back(st[0].back());
        st[0].pop_back();
    }
    while((int)st[1].size() > (n+1)/3) {
        st[2].push_back(st[1].back());
        st[1].pop_back();
    }
    vector<int> p(n);
    rep(col, 3){
        rep(i, st[col].size()){
            p[st[col][i]] = i*3 + col + 1;
        }
    }
    rep(i,n){
        cout << p[i] << (i < n-1 ? " " : "\n");
    }
    return 0;
}