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;

#ifdef LOCAL
#define debug(var) cerr << #var << " = " << var << endl;
#else
#define debug(var)
#endif

void init() {
	ios_base::sync_with_stdio(0);
	cin.tie(0); cout.tie(0);
}

const int N = 2e5+345;

vector<int> g[N];
int ans[N];

int bio[N];
void dfs(int u, int c=1) {
	bio[u] = c;
	for (int v: g[u]) {
		if (bio[v]) continue;
		dfs(v, c==1?2:1);
	}
}

void solve() {
	int n; scanf("%d", &n);
	for (int i = 0; i < n-1; ++i) {
		int a, b; scanf("%d%d", &a, &b);
		g[--a].push_back(--b);
		g[b].push_back(a);
	}
	dfs(0);
	int r = 0;
	int c = 1;
	for (int i = 0; i < n; ++i) if (bio[i] == c) ++r;
	if (r > n-r) {
		r = n-r;
		c = 2;
	}
	debug(r);
	if (r <= n/3) {
		int k = 1;
		for (int i = 0; i < n; ++i) {
			if (bio[i] == c && k*3 <= n) {
				ans[i] = k*3; ++k;
			}
		}
	} else {
		int k = 0, q = 0;
		for (int i = 0; i < n; ++i) {
			if (bio[i] == c) {
				if (k*3+1 <= n) {
					ans[i] = k*3+1;
					++k;
				}
			} else {
				if (q*3+2 <= n) {
					ans[i] = q*3+2;
					++q;
				}
			}
		}
	}
	set<int> s;
	for (int i = 1; i <= n; ++i) s.insert(i);
	for (int i = 0; i < n; ++i) {
		if (ans[i]) s.erase(ans[i]);
	}
	for (int i = 0; i < n; ++i) {
		if (!ans[i]) {
			ans[i] = *s.begin();
			s.erase(s.begin());
		}
		printf("%d ", ans[i]);
	}
}

int main() {
	init();
	int t = 1; //scanf("%d", &t);
	while (t--) {
		solve();
	}
	return 0;
}