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;
#define pb push_back
const int N = 2e5 + 5, INF = 0x3f3f3f3f;
inline int read() {
	int x = 0, f = 0; char ch = 0;
	while (!isdigit(ch)) f |= ch == '-', ch = getchar();
	while (isdigit(ch)) x = (x << 3) + (x << 1) + (ch ^ 48), ch = getchar();
	return f ? -x : x;
}
int ans[N];
vector <int> G[N], col[2], mod[3];
void dfs(int x, int c, int fa) {
	col[c].push_back(x);
	for (auto y : G[x]) if (y != fa) dfs(y, c ^ 1, x);
}
void sol(int i, int j) {
	while (!col[i].empty() && !mod[j].empty()) {
		ans[col[i].back()] = mod[j].back();
		col[i].pop_back(), mod[j].pop_back();
	}
}
int main() {
	int n = read();
	for (int i = 1; i < n; i++) {
		int x = read(), y = read();
		G[x].pb(y), G[y].pb(x);
	}
	for (int i = 1; i <= n; i++) mod[i % 3].pb(i);
	dfs(1, 0, 0);
	sol(col[0].size() < col[1].size(), 1), sol(col[0].size() < col[1].size(), 2);
	sol(0, 0), sol(1, 0);
	for (int i = 1; i <= n; i++) printf("%d ", ans[i]);
	return 0;
}