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;
typedef long long ll;
typedef vector<ll> vi;
typedef pair<ll,ll> pi;
typedef vector<pi> vpi;
typedef long double ld;
#define pb emplace_back
#define mp make_pair
#define lb lower_bound
#define ub upper_bound
#define ALL(x) x.begin(), x.end()
#define SZ(x) (ll)x.size()
#define f first
#define s second
#define MAXN 400010
const ll MOD = 998244353;
const ll INF = 1e15;

ll N;
vi V[MAXN];
ll a,b;
ll dst[MAXN];
vi O;
vi E;
ll out[MAXN];

void dfs(int x, int p){
	for (auto v:V[x])if(v!=p){
		dst[v] = dst[x]+1;
		dfs(v,x);
	}
}

ll done[MAXN];

int main(){
	cin>>N;
	for (int i=0;i<=N;++i){
		cin>>a>>b;
		V[a].pb(b);
		V[b].pb(a);
	}
	dfs(1,-1);

	for (int i=1;i<=N;++i){
		if (dst[i]%2==1)O.pb(i);
		else E.pb(i);
	}
	if (SZ(O) > SZ(E))swap(O,E);

	ll h = N/3;

	if (SZ(O) > h){
		ll t=1;
		while (SZ(O) && t <= N){
			out[O.back()] = t;
			done[t] = 1;
			O.pop_back();
			t += 3;
		}
	}
	if (SZ(O) > h){
		ll t=2;
		while (SZ(O) && t <= N){
			out[O.back()] = t;
			done[t] = 1;
			O.pop_back();
			t += 3;
		}
	}
	ll t=3;
	while (SZ(O) && t <= N){
		out[O.back()] = t;
		done[t] = 1;
		O.pop_back();
		t += 3;
	}
	vi rem;
	for (int i=1;i<=N;++i)if (!done[i])rem.pb(i);
	for (int i=1;i<=N;++i){
		if (out[i] == 0){out[i] = rem.back();rem.pop_back();}
	}
	for (int i=1;i<=N;++i)cout<<out[i]<<' ';
}