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;
constexpr int maxn = 200000 + 1;
vector<int> G[maxn];
int d[maxn], p[maxn];
stack<int> A[2];
stack<int> B[3];
void DFS(int u, int par = 0){
    A[d[u]].push(u);
    for(int v : G[u]) if(v != par){
        d[v] = d[u] ^ 1;
        DFS(v, u);
    }
}
void fuck(stack<int>& A, stack<int>& B){
    while(not A.empty() and not B.empty()){
        p[A.top()] = B.top();
        A.pop();
        B.pop();
    }
}
int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    int N;
    cin >> N;
    for(int i = 1, u, v; i < N; i += 1){
        cin >> u >> v;
        G[u].push_back(v);
        G[v].push_back(u);
    }
    DFS(1);
    for(int i = 1; i <= N; i += 1) B[(i + 2) % 3].push(i);
    if(A[0].size() >= B[0].size() and A[1].size() >= B[1].size()){
        fuck(A[0], B[0]);
        fuck(A[1], B[1]);
        fuck(A[0], B[2]);
        fuck(A[1], B[2]);
    }
    else if(A[0].size() >= B[1].size() and A[1].size() >= B[0].size()){    
        fuck(A[1], B[0]);
        fuck(A[0], B[1]);
        fuck(A[0], B[2]);
        fuck(A[1], B[2]);
    }
    else if(B[2].size() >= A[0].size()){
        fuck(A[0], B[2]);
        fuck(A[1], B[0]);
        fuck(A[1], B[1]);
        fuck(A[1], B[2]);
    }
    else if(B[2].size() >= A[1].size()){
        fuck(A[1], B[2]);
        fuck(A[0], B[0]);
        fuck(A[0], B[1]);
        fuck(A[0], B[2]);
    }
    else assert(0);
    for(int i = 1; i <= N; i += 1) cout << p[i] << " ";
    return 0;
}