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;

const int N = 200005;
int n, a[N], ans[N];
vector<int> e[N], v[2];

void dfs(int x, int fa, int d) {
    v[d].push_back(x);
    for (int to : e[x])
        if (to != fa)
            dfs(to, x, d ^ 1);
}

int main() {
    scanf("%d", &n);
    for (int i = 1; i < n; i++) {
        int x, y;
        scanf("%d%d", &x, &y);
        e[x].push_back(y);
        e[y].push_back(x);
    }
    dfs(1, 0, 0);
    int k1 = n / 3, k2 = n / 3;
    if (n % 3 == 1)
        k1++;
    else if (n % 3 == 2)
        k1++, k2++;
    if (v[0].size() > v[1].size())
        swap(v[0], v[1]);
    if (k1 <= v[0].size() && k2 <= v[1].size()) {
        for (int i = 0; i < k1; i++)
            ans[v[0][i]] = 1;
        for (int i = 0; i < k2; i++)
            ans[v[1][i]] = 2;
    } else {
        for (int i = 0; i < k1; i++)
            ans[v[1][i]] = 1;
        for (int i = k1; i < k1 + k2; i++)
            ans[v[1][i]] = 2;
    }
    int now0 = 3, now1 = 1, now2 = 2;
    for (int i = 1; i <= n; i++) {
        if (!ans[i])
            a[i] = now0, now0 += 3;
        else if (ans[i] == 1)
            a[i] = now1, now1 += 3;
        else
            a[i] = now2, now2 += 3;
    }
    for (int i = 1; i <= n; i++){
        printf("%d ", a[i]);
    }
}