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 int long
using pi=pair<int,int>;
struct edge{
    int a, b;
    edge(){};
    edge(int x, int y)
        : a(x)
        , b(y){};
};
signed main() {
    int n;
    cin>>n;
    vector<edge> E(n - 1);
    vector<vector<int>> G(n);
    for (int i = 0; i < n - 1; ++i) {
        cin >> E[i].a >> E[i].b;
        --E[i].a, --E[i].b;
        G[E[i].a].push_back(E[i].b);
        G[E[i].b].push_back(E[i].a);
    }
    int m0 = 3, m1 = 1, m2 = 2;
    queue<pi> que;
    vector<bool> vis(n, false);
    vector<int> col(n, 0),res(n);
    int cred = 0, cblue = 0;
    que.push(pi(0, -1));
    while (!que.empty()) {
        pi q = que.front();
        que.pop();
        col[q.first] = q.second;
        ++(q.second==-1 ? cred : cblue);
        vis[q.first] = true;
        for (auto i : G[q.first]) {
            if (!vis[i]) que.push(pi(i, -3-q.second));
        }
    };

    for(int i=0;i<n;++i){
        if(m2<=n&&(cred*3<=n&&col[i]==-2&&m1>m2 || cblue*3<=n&&col[i]==-1&&m1>m2 || min(cred,cblue)*3>n&&col[i]==-2)){
            res[i] = m2;
            m2 += 3;
        }
        else if(m1<=n&&(cred*3<=n&&col[i]==-2&&m1<=m2 || cblue*3<=n&&col[i]==-1&&m1<=m2 || min(cred,cblue)*3>n&&col[i]==-1)){
            res[i] = m1;
            m1 += 3;
        }
        else{
            res[i]=m0;
            m0 += 3;
        }
    }

    for (int i = 0; i < n; ++i) cout << (i ? " " : "") << res[i];
    cout << endl;
}