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;
int const nax = 2e5 + 10;

vector<int> a[nax], node[2];

int n , ct[3], c[3], col[nax];
int  ptr1 = -2, ptr2 = -1, ptr3  = 0;

void dfs (int nod,  int pa, int lv) {
    ct[lv % 2] ++;
    node[lv % 2].push_back(nod);
    for (int i : a[nod])
        if (i != pa)
            dfs(i, nod, lv + 1);
}

int main () {
    scanf("%d", &n);
    for (int i = 1 ; i < n ; ++ i) {
        int u, v;
        scanf("%d %d", &u, &v);
        a[u].push_back(v);
        a[v].push_back(u);
    }

    c[0] = n / 3 ;
    c[1] = n / 3 + (n % 3 > 0);
    c[2] = n / 3 + (n % 3 > 1);

    dfs(1, 0, 0);

    if (max(ct[0], ct[1]) >= c[1] + c[2]) {
        int toCol = ct[0] >= ct[1] ? 0 : 1;
        for (int i = 0 ;i < c[1] ; ++ i)
            col[node[toCol][i]] = (ptr1 += 3);

        for (int i = 0 ; i < c[2] ; ++ i)
            col[node[toCol][i + c[1]]] = (ptr2 += 3);
    }
    else if (max(ct[0], ct[1]) >= c[1] && min(ct[0], ct[1]) >= c[2]) {
        int maxCol = ct[0] >= ct[1] ? 0 : 1;
        int minCol = (maxCol ^ 1);
        for (int i = 0 ; i < c[1] ; ++ i)
            col[node[maxCol][i]] = (ptr1 += 3);

        for (int i = 0 ; i < c[2] ; ++ i)
            col[node[minCol][i]] = (ptr2 += 3);
    }

    for (int i = 1 ; i <= n ; ++ i)
        printf("%d ", col[i] ? col[i] : ptr3 += 3);


}
/*
    Good Luck
        -Lucina
*/