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 <cstdio>
#include <algorithm>
#include <cctype>
#include <cstring>
#include <vector>
using namespace std;
inline int read(int f = 1, int x = 0, char ch = ' ')
{
    while(!isdigit(ch = getchar())) if(ch == '-') f = -1;
    while(isdigit(ch)) x = x*10+ch-'0', ch = getchar();
    return f*x;
}
const int N = 2e5+5;
int n, k, c[2], f[3], a[3][N], d[N], p[N]; vector<int> e[N];
void dfs(int x, int fa)
{ ++c[d[x]]; for(auto y:e[x]) if(y != fa) d[y] = d[x]^1, dfs(y, x); }
int main()
{
    n = read(); for(int i = 1, x, y; i < n; ++i) x = read(), y = read(), e[x].push_back(y), e[y].push_back(x);
    dfs(1, 0), k = c[0]<c[1]?0:1;
    for(int i = 1; i <= n; ++i) a[i%3][(i-1)/3+1] = i;
    if(c[k] <= n/3)
    {
        for(int i = 1; i <= n; ++i)
            if(d[i] == k) p[i] = a[0][++f[0]];
            else if(f[2] < (n+1)/3) p[i] = a[2][++f[2]];
            else if(f[1] < (n+2)/3) p[i] = a[1][++f[1]];
            else p[i] = a[0][++f[0]];
    }
    else 
    {
        for(int i = 1; i <= n; ++i)
            if(d[i] == k) 
            {
                if(f[2] < (n+1)/3) p[i] = a[2][++f[2]];
                else p[i] = a[0][++f[0]];
            }
            else if(f[1] < (n+2)/3) p[i] = a[1][++f[1]];
            else p[i] = a[0][++f[0]];     
    }
    for(int i = 1; i <= n; ++i) printf("%d ", p[i]); puts("");
    return 0;
}